Astronomy

Problem about finding distance with magnitude given

Problem about finding distance with magnitude given

A star in Pleiades cluster has apparent magnitude of +12 ($m_{star} = +12$) determine it's distance from the sun? (We know the Sun properties such apparent and absolute magnitudes, distance from earth, luminosity, brightness (energy flux) and others but nothing is given about the other star expect that apparent magnitude).

My problem is that we don't have luminosity of the star. We can say:

$m_{sun} - m_{star} = -2.5 * log_{10}{frac{b_{sun}} {b_{star}}} Rightarrow 10^{(-26.8 - 12)/(-2.5)} approx 3.31131 * 10^{15} = frac{b_{sun}} {b_{star}} approx frac{1370 Wm^-2}{L_{star} / (4 pi d^2)} $

As we can see, we don't have luminosity of the star. Can this question be solved with this information?


Short answer: no, you can't solve the question if all you know is its apparent magnitude. It could be a faint star that's nearby, or a very luminous star that's far away. (The Sun's parameters are irrelevant.)

(I'm ignoring the bit about it being "a star in the Pleiades", which implies that it's a physical member of the Pleiades cluster, in which you already know its distance: it's the same as the distance of the Pleiades cluster. Maybe you just meant "it's in the same part of the sky".)


Problem about finding distance with magnitude given - Astronomy

Fomalhaut, or Alpha Pisces Austrinis, as seen from Stellarium software.

One of the fundamental requirements for astronomy to work as a science is the need to accurately measure the distances to objects. Astronomers have developed a series of methods for measuring stellar and cosmic distances, which fit together and inform each other. Taken together, they’re referred to as the Distance Ladder, as each one provides the basis for the next step out in distance.

Website for our MIT BLOSSOMS video lesson plan on using the parallax method to find the distances to nearby stars.

For the first rung on the Distance Ladder, for objects closer than about 100 light years, we use the parallax method. I developed a lesson plan to teach this and my students and I made a video of this lesson for the MIT BLOSSOMS project. You can go to their website to see this video and download the lesson plan at:

I’ve also written a previous post about the parallax method here:

As I developed lesson plans to use on my poster for the American Astronomical Society conference, I decided to revise my lesson that uses absolute and apparent magnitudes to determine the distances to stars. It would be a good way to introduce collecting and using astronomical data. My students were already familiar with the constellations, stellar classifications, and the Hertzsprung-Russell Diagram, they had been gathering data about the stars using Stellarium software, and had completed the parallax lesson, so they were ready to go. But this lesson requires some explanation:

Stellar Magnitudes:

Beyond 100 or so light years, we have to use a method called the Distance Modulus formula. To use it, one has to measure how bright a star appears (apparent magnitude, or m) with high accuracy. How bright a star appears is based on two things: how close it is and how much light the star actually emits. Astronomers eliminate differences in distance as a factor by pretending to move all stars to same distance: 32.6 light years or ten parsecs. Their brightness from this distance is called their absolute magnitude (M).

Separating out a spectrum of the star’s wavelengths provides a fingerprint that identifies the star’s classification, using the pattern of absorption lines and flux densities based on Wien’s Law. We’ve learned enough about each type and sub-type of star to know how much total light it gives off. This is called its luminosity, and is measured compared to our sun.

Hipparchus, who created the first accurate star catalog about 130 B.C. He provided magnitude numbers from 1 for the brightest stars to 6 for the dimmest that could be seen with the unaided eye.

Hipparchus, Herschel, and Newton:

To figure out the distance to a star is therefore to compare how bright a star really is (absolute magnitude) with how bright a star appears (apparent magnitude). Now this isn’t quite as straightforward as it seems. Sir Isaac Newton discovered that light follows an inverse square law – that the brightness of a light falls with the square of its distance. In other words, a light that is twice as far away will be one fourth as bright as before. It is an exponential curve.

Greek astronomers, as painted by Raphael in The School of Athens. Hipparchus is holding the celestial sphere.

Another problem is that the original magnitude scale was developed by the Greek astronomer Hipparchus in about 150 B.C. He created the first star catalog and assigned the stars numbers based on their perceived brightness, with the brightest star (Sirius) given the number 1 and the dimmest star visible the number 6. This inverted scale has stuck with us and can be a bit tricky to understand. The important thing is that the higher the magnitude number, the dimmer the star is. Lower numbers mean brighter stars.

Sir William Herschel, who discovered that five differences in magnitude are about 100 times difference in brightness.

Once Newton put light on a mathematical basis, astronomers wanted to standardize the magnitude system so that mathematical formulas could be used. William Herschel, along with his sister Caroline, cataloging thousands of stars (and discovered Uranus along the way). They discovered that a magnitude 1 star was roughly 100 times brighter than a magnitude 6 star, or that five orders of magnitude produce a 100 fold change in brightness. Using this, the magnitude scale was adjusted to make it come out exactly 100 times, so some stars such as Sirius now have negative apparent magnitudes.

The Modulus Formula:

With the magnitude scale adjusted to fit a logarithmic curve, you can now say that one star is exactly so many times brighter than another. You can represent this relationship with the formula: (M – m – 5)/-5 = logD , where M is the absolute magnitude of the star, m is the apparent magnitude, and D is the distance in parsecs.

Let’s work this through for the star Fomalhaut. It has an apparent magnitude of 1.15, and an absolute magnitude of 1.72. So plugging in the numbers gives us: (1.72 – 1.15 – 5)/-5 = 0.886 = logD. Taking the antilog of 0.886 gives us 7.69 parsecs. Since there are 3.26 light years in a parsec, the distance to Fomalhaut is therefore 25.1 light years. Now, since this is also a nearby star, we can use the parallax method to double check the distance. The methods in the Distance Ladder back each other up.

Modulus Method Lesson Page 1

It gets harder when stars are so far away that you can’t accurately measure their apparent magnitude, or if they are in distant galaxies, etc. There are other methods in the Distance Ladder, such as Hubble’s Law, that can give a distance in an expanding universe based on the degree of a galaxy’s red shift. In between, there is a method first pioneered by Henrietta Leavitt based on the precise period-luminosity function of Cepheid variable stars. Hubble used her work to determine the distance to the Andromeda galaxy.

The Magnitude – Luminosity Function. Since luminosity varies with the inverse of the distance squared, it is an exponential curve. In this case, if you know the ratio of luminosity between two stars, you can use the curve to determine the differences in magnitudes. The modulus formula uses logarithms to do the same calculations.

The Lesson Plan:

Now that I’ve explained all that, its time to try out the lesson plan. All three pages are attached here and can be downloaded. I had an older version of it in the form of often duplicated printouts, but no remaining digital copy, so I scanned the pages in and revised the explanation and data tables to make the process work better, then put student samples on my poster for AAS.

In addition to using the Modulus Formula directly, the lesson shows how to do the same using the inverse square light curve. With the curve, one can figure out the magnitude differences (between apparent and absolute or between two different stars, such as Fomalhaut and Alpha Centauri) and determine the difference in brightness. Or go the other direction – knowing the differences in brightness, one can determine the differences in magnitudes.

Modulus lesson plan Page 3

One trick is remembering whether a star has to be moved forward or back to get it to 32.6 light years. If forward, its absolute magnitude will be a lower number than its apparent magnitude. If the star is closer than 32.6 light years, then the apparent magnitude will be less that the absolute magnitude (it appears brighter because it is close to us – remember that it is a reversed scale).

After we did the parallax simulation activity together as a class, I had my students learn the Modulus method. I then taught them how to make the representative color images using WISE data presented in my last post. Most of them were able to grasp the concepts well, based on my feedback quizzes and assessment forms.

I’ve made a few more modifications for the purpose of posting the pages here, such as adding extra columns on the first page to facilitate doing more calculations. Eventually I need to make brand new digital versions that are easier to edit. Although effective, I’m still not quite satisfied with the flow or layout of the lesson.


Magnitude & Direction of a Vector

A Vector is something that has two and only two defining characteristics.

Trait #1) Magnitude

Trait #2) Direction

Examples of Vectors Non Examples
4 units long at 30 $^$ 4 unit
44 miles per hour east (velocity) speed of 44 mph (speed)

Practice Problems

Problem 1

Describe using compass directions (North, South, East, West) the direction of the vector pictured below.

Problem 2

What is the magnitude and the direction of the vector below?

Problem 3

Find the magnitude and direction of vector in the diagram below.

The direction of the vector is 47° North of West, and the vector's magnitude is 2.

Problem 4

Find the magnitude and direction of vector in the diagram below.

The direction of the vector is 43° East of South, and the vector's magnitude is 3. It is also possible to describe this vector's direction as 47. South of East.

Problem 5

What is the difference between a vector that is 55° north of west and a vector that's 35° west of north?

Show Answer

In terms of direction, there is no difference whatsoever between 55° north of west, and 35° west of north.


Absolute Value-Magnitude and Distance

Examples, videos, and solutions to help Grade 6 students understand the absolute value of a number as its distance from zero on the number line.

Students use absolute value to find the magnitude of a positive or negative quantity in a real-world situation.

New York State Common Core Math Grade 6, Module 3, Lesson 11

Opening Exercises
What is the relationship between the following pairs of numbers? How do each pair of numbers relate to zero?
-4 and 4
-2 1/2 and 2 1/2
-10 and 10

What is the absolute value of a number?

The absolute value of a number is the distance between the number and zero on a number line. Every number and it opposite are the same distance from zero on the number line.In other words, a number and its opposite have the same absolute value.

What is the absolute value of 6?
What is the absolute value of -6?
Both 6 and -6 are six units from zero.
What is the absolute value of 0?

Example 1: The Absolute Value of a Number

The absolute value of ten is written as |10|. On the number line, count the number of units from 10 to 0. How many units is 10 from 0?
What other number has an absolute value of ? Why?

The absolute value of a number is the distance between the number and zero on the number line.

Example 2: Using Absolute Value to Find Magnitude

Mrs. Owens received a call from her bank because she had a checkbook balance of dollars. What was the magnitude of the amount overdrawn?

The magnitude of a quantity is found by taking the absolute value of its numerical part

4. Maria was sick with the flu and her weight change as a result of it is represented by -4 pounds. How much weight did Maria lose?

5. Jeffrey owes his friend $5. How much is Jeffrey&rsquos debt?

6. The elevation of Niagara Falls, which is located between Lake Erie and Lake Ontario, is 326 feet. How far is this above sea level?

7. How far below zero is -16 degrees Celsius?

8. Frank received a monthly statement for his college savings account. It listed a deposit of $100 as +100.00. It listed a withdrawal of $25 as -25.00. The statement showed an overall ending balance of $835.50. How much money did Frank add to his account that month? How much did he take out? What is the total amount Frank has saved for college?

9. Meg is playing a card game with her friend Iona. The cards have positive and negative numbers printed on them. Meg exclaims: &ldquoThe absolute value of the number on my card equals 8!&rdquo What is the number on Meg&rsquos card?

10. List a positive and negative number whose absolute value is greater than . Explain how to justify your answer using the number line.

11. Which of the following situations can be represented by the absolute value of10? Check all that apply.
The temperature is degrees below zero. Express this as an integer.
Determine the size of Harold&rsquos debt if he owes .
Determine how far is from zero on a number line.
degrees is how many degrees above zero?

12. Julia used absolute value to find the distance between 0 and 6 on a number line. She then wrote a similar statement to represent the distance between and . Below is her work. Is it correct? Explain.
|6| = 6, |-6| = -6

13. Use absolute value to represent the amount, in dollars, of a $238.25 profit.

14. Judy lost 15 pounds. Use absolute value to represent the number of pounds Judy lost.

15. In math class, Carl and Angela are debating about integers and absolute value. Carl said two integers can have the same absolute value and Angela said one integer can have two absolute values. Who is right? Defend your answer.

16. Jamie told his math teacher: &ldquoGive me any absolute value, and I can tell you two numbers that have that absolute value.&rdquo Is Jamie correct? For any given absolute value, will there always be two numbers that have that absolute value?

17. Use a number line to show why a number and its opposite have the same absolute value.

18. A bank teller assisted two customers with transactions. One customer made a $25.00 withdrawal from a savings account. The other customer made a $15 deposit. Use absolute value to show the size of each transaction. Which transaction involved more money?

19. Which is farther from zero: -7 3/4 or 7 1/2? Use absolute value to defend your answer.

I am thinking of two numbers. Both numbers have the same absolute value. What must be true about the two numbers?

Can the absolute value of a number ever be a negative number? Why or why not?

Problem Set
For each of the following two quantities in Problems 1–4, which has the greater magnitude? (Use absolute value to defend your answers.)
1. 33 dollars and -52 dollars
2. -14 feet and 23 feet
3. -24.6 pounds and -24.58 pounds
4. -11 1/4 degrees and 11 degrees

Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.


Vector Magnitude

In this lesson, we will learn how to find the magnitude of 2-dimensional vectors and 3-dimensional vectors.



The length of a vector is called the magnitude or modulus of the vector.

The following diagram shows the magnitude of a vector. Scroll down the page for more examples and solutions to calculate the magnitude of 2-D and 3-D vectors .

Express each of the following vectors as a column vector and find its magnitude.

Vectors in 2D

Vectors in 3D

Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.


Displacement and distance problems with solutions

Problems and Solutions about distance and displacement are presented and updated useful for high school and college students.

An object moves from point A to B, C, and D finally, along a rectangle.
(a) Find the magnitude and direction of the displacement vector of the object?
(b) Find the distance traveled by that object?
(c) Suppose the object returns to point A, its initial position. Now, Find the displacement and distance?

(b) distance is equal to the length of AB+BC+CD.
(c) Since the initial and final positions are the same, so by definition of displacement, the difference between them is zero. But distance, in this case, is the perimeter of the rectangle.

An object moves along a right triangle from point A to B to C shown in the figure below. (Consider the sides as $3,< m m>$ and $4,< m m>$)
(a) Find the magnitude and direction of the displacement vector?
(b) How much distance traveled by this moving object?
(c) Suppose the object returns to point A, its initial position. Now, Find the displacement and distance?

Solution (2):
(a) Connect the initial (A) and final (C) points together by the straightest line. The magnitude and direction of that line represent the displacement vector. Here, the displacement is the hypotenuse of the right triangle. So using the Pythagorean theorem, one finds its magnitude as egin c^2 &= a^ <2>+ b^ <2> Rightarrow c&=sqrt +b^<2>> &= sqrt <3^<2>+ 4^<2>>=5, < m m>end the direction of displacement is shown in figure (to the northeast).

(b) Adding the base and height of this triangle gets the distance traveled from (A) to (C).
egin ext &= a + b &= 3 + 4 &= 7, < m m>end

(c) Displacement is zero since the object returns to the initial position. Distance is also computed as the perimeter of the triangle.

Problem (3):
A displacement vector has a magnitude of $810,< m m>$, and points at an angle of $18^circ$ above the positive x-axis. What are the $x$ and $y$ components of this vector?

Solution (3):
Given the length and direction of a vector such as displacement $vec$ , one can find its components by the following formulas egin R_x &= |vec|cos heta &=(810)cos 18^circ=770.35,< m m> ext R_y&=|vec|sin heta &=(810)sin 18^circ=250.30, < m m>end

Problem (4):
A displacement vector is $23,< m km>$ in length and directed $65^circ$ south of east. What are the components of this vector?

Solution (4):
Like previous problem, displacement in the components form is as $vec=|vec|,cos heta hat+|vec|,sin heta hat$ where $ heta$ is measured with the positive $x$-axis. Here, the given angle is below the $x$-axis so we have
egin R_x &= |vec|cos heta &=(23)cos (-65^circ)=9.72,< m m> ext R_y&=|vec|sin heta &=(23)sin (-65^circ)=-20.84, < m m>end The minus sign of the y-component indicates the south direction.

Problem (5):
A race car will make one lap around a circular track of radius $R$. When the car has traveled halfway around the track, what is the magnitude of displacement of the car from the starting point? Find the distance of the car?

Solution (5):
Displacement is the difference between initial and final points so in this case, that difference is the diameter of the circular track. Thus,
[ ext=2R]
Distance is also the circumference of the semi-circle that is $2,pi R$.

Problem (6):
You walk $5,< m km>$ due south then $4,< m km>$ due west. What is the magnitude of your displacement?

Solution (6):
$5,< m km>$ south means a position vector of magnitude $5,< m km>$ from the origin toward the south. The other journey starts from the tail of the previous vector with magnitude and direction of $4,< m km>$ and west, respectively.

In the mathematical language the above two vectors are written as $vec_1=5,(-hat)$ and $vec_2=4,(-hat)$. To find the displacement we have two options, either using the drawing vector or the algebraic method which is efficient for more involved problems.

Method I:
The vector that constructed from connecting the tip of the first vector to the tail of the second vector is displacement (triangle method). Use the Pythagorean theorem to get the magnitude of the desired vector.


Method II:
In the case of several journey segments, displacement can be determined using the methods of vector addition. Here, the whole journey is formed of the two segments with the corresponding vectors $vec_1=5,(-hat)$ and $vec_2=4,(-hat)$. Thus, using the vector addition method we get
egin vec&=vec_1+vec_2 &=5,(-hat)+4,(-hat) ext &=-4,hat-5,hat end The magnitude and direction of the above vector is determined by the following formulas
egin |vec|&=sqrt &=sqrt<(-5)^2 + (-4)^2> &=sqrt<41>,< m km> ext heta &= an^<-1>left(frac ight) &= an^<-1>left(frac<-5><-4> ight) &=51.34^circ end Since the $x$ and $y$ components of the displacement place in the third quadrant so the above angle is toward the west south.

Important notes about displacement and distance:

  • If the initial and final points are the same, so the total displacement is zero.
  • Distance is measured as the perimeter of the path traveled by the moving object.

For more involved and difficult problems with a thorough explanation about the definition of displacement in two and three dimensions visit the exam center or course pages .

In addition, you can also check out the Wikipedia article about distance and displacement .


Magnitude and Direction of Vectors

The magnitude of a vector P Q &rarr is the distance between the initial point P and the end point Q . In symbols the magnitude of P Q &rarr is written as | &thinsp P Q &rarr &thinsp | .

If the coordinates of the initial point and the end point of a vector is given, the Distance Formula can be used to find its magnitude.

| &thinsp P Q &rarr &thinsp | = ( x 2 &minus x 1 ) 2 + ( y 2 &minus y 1 ) 2

Find the magnitude of the vector P Q &rarr whose initial point P is at ( 1 , 1 ) and end point is at Q is at ( 5 , 3 ) .

Substitute the values of x 1 , y 1 , x 2 , and y 2 .

The magnitude of P Q &rarr is about 4.5 .

Direction of a Vector

The direction of a vector is the measure of the angle it makes with a horizontal line .

One of the following formulas can be used to find the direction of a vector:

tan &theta = y x , where x is the horizontal change and y is the vertical change

tan &theta = y 2 &thinsp &minus &thinsp y 1 x 2 &thinsp &minus &thinsp x 1 , where ( x 1 , y 1 ) is the initial point and ( x 2 , y 2 ) is the terminal point.

Find the direction of the vector P Q &rarr whose initial point P is at ( 2 , 3 ) and end point is at Q is at ( 5 , 8 ) .

The coordinates of the initial point and the terminal point are given. Substitute them in the formula tan &theta = y 2 &thinsp &minus &thinsp y 1 x 2 &thinsp &minus &thinsp x 1 .

Find the inverse tan, then use a calculator.

The vector P Q &rarr has a direction of about 59 ° .

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How do I get stellarium to show a comet's apparent magnitude

I've been tracking 46p/ Wirtanen in Stellarium but it shows the magnitude being 8 or 9 even when it was supposed to be 4 on that one day. It is just labeled as Magnitude. But for the rest of the objects, it shows the actual apparent magnitude. So how do I get it to do that for Wirtanen?

#2 gzotti

It seems the magnitude entry from the MPC elements was a bit off.

Solution: Edit data/ssystem_minor.ini.

Details: Appendix D2.2 in the User Guide.

#3 Alexander Wolf

I've been tracking 46p/ Wirtanen in Stellarium but it shows the magnitude being 8 or 9 even when it was supposed to be 4 on that one day. It is just labeled as Magnitude. But for the rest of the objects, it shows the actual apparent magnitude. So how do I get it to do that for Wirtanen?

#4 AstronomicalBirb

If you know good model for predicting the magnitudes of comets - please suggest us it or give link - it maybe very helpful for all astronomy amateurs and developers of planetariums.

Oh I know nothing at all about it. I'm just asking basically, why is the magnitude off. The Apparent magnitude should be around 5 now though in stellarium it is almost 10. So I am assuming that it is not showing the apparent magnitude but the absolute magnitude which is around 15 or so.

#5 AstronomicalBirb

It seems the magnitude entry from the MPC elements was a bit off.

Solution: Edit data/ssystem_minor.ini.

Details: Appendix D2.2 in the User Guide.

Can you give more detail on how I can fix this?

#6 sg6

Suspect it is not possible with any degree of accuracy.

The comet will out gas and shed material and that is what we see as the magnitude. If for whatever reason more material is evaporated off it is "brighter" both as illuminated material and as size. If however very little is shed/evaporated then it remains dim.

Any that have it "accurate" is I would say more down to luck and some estimate someone performed.

If Stellarium were to do a daily update then we would have to down load the appropriatre data in effect daily download of the data. Stellarium is not an interactive web site.

Remember Ison, predicted to be the brightest comet of the century. That was a bit of a non-event at the end. It did not brighten as they had expected.

Solar activity will effect the comet brightness. More solar wind, more stuff knocked off, more atoms excited and emitting light, more reflected light.

#7 Jon Isaacs

It seems the magnitude entry from the MPC elements was a bit off.

Solution: Edit data/ssystem_minor.ini.

Details: Appendix D2.2 in the User Guide.

This seems to be an issue with all the planetarium software. About the 13th of December, suddenly the magnitude of 46P went from 3.8 to 8.8. I noticed it in Sky Safari, tried a few other programs and they had the same issue. I eventually contacted someone at the MPC and tried to discuss the issue. At the time, 46 P was naked eye from relatively dark skies. Obviously it was not magnitude 8.8..

The guy did respond but basically blew me off.

#8 gzotti

Can you give more detail on how I can fix this?

Follow instructions to edit ssystem_minor.ini

Find entry for 46P/Wirtanen

Change absolute_magnitude to 10 or so.

Edited by gzotti, 01 January 2019 - 05:54 AM.

#9 gzotti

This seems to be an issue with all the planetarium software. About the 13th of December, suddenly the magnitude of 46P went from 3.8 to 8.8. I noticed it in Sky Safari, tried a few other programs and they had the same issue. I eventually contacted someone at the MPC and tried to discuss the issue. At the time, 46 P was naked eye from relatively dark skies. Obviously it was not magnitude 8.8..

The guy did respond but basically blew me off.

Jon

I assume MPC does not comment issues in other people's software.

As one original issue I would see the brightness development of this comet. See http://www.aerith.ne. 0046P/2018.html

See the apparent brightness surge around October or so (-4 magnitudes?). MPC gives absolute magnitude as 14. Still today. Edit ssystem_ini and set this value to e.g.9 or 10.

#10 Jon Isaacs

Everyone's software that use MPC data seems to be giving the same result.

#11 gzotti

Computing comet brightness requires an algorithm (formula) and data. There is a standard formula which has been used for decades, and is quite probably used by all software titles out there. And there is the data, usually provided in high quality by the MPC. And there is the operator (observer) who can judge when data may be off.

I cannot reproduce a sudden brightness drop from 3.8 to 8.8 when using the same set of data elements. I see a magnitude constantly about 4 or 5 mag too dim, based on a data entry of absolute_magnitude=14 (taken from the MPC data). Manually editing this value to 10 or 9 seems to provide a better overall brightness development.

#12 kb7wox

The guy did respond but basically blew me off.

Wow! C/1973 E1 still casting a long shadow

#13 catalogman

If you know good model for predicting the magnitudes of comets - please suggest us it or give link - it maybe very helpful for all astronomy amateurs and developers of planetariums.

The general formula for the magnitude H of a comet is

H0 = absolute magnitude (magnitude at d = r = 1 A.U.)

n = 2: reflection from a constant amount of gas and dust (ideal case)

n = 4: gas emanating from nucleus, varies with heliocentric distance (most common case)

n = 6: more gas emanating from nucleus, varies with heliocentric distance (extreme case)

[Brandt John C., Hodge Paul W. Solar System Astrophysics (McGraw-Hill, 1964), p. 222]

For Stellarium, a solution is to add an option to let the observer calibrate the value of n for
a given date instead of using the slope parameter as a predictor. For instance, if a magazine article
gives a comet magnitude of 11.8 on January 1, the observer inputs that date and then adjusts the value
of n with a slide bar until H=11.8. This revised value of n is then used for later predictions.

Another more complicated solution would require more observations and the observer's aperture:

Edited by catalogman, 06 January 2019 - 05:45 PM.

#14 gzotti

Yes, that's the canonical formula that is also used in Stellarium. And thanks for the insight into the meaning of the slope parameter.

Now, e.g. see Dec. 5, 2018:
d=0.1AU. Log d=-1.

0.025 which is pretty small. This second factor in this distance only has a small influence near r=1.
If we take MPC's H0=14 for real we have

With a reported estimate of H=5.5 this reveals: (numerically don't believe it!)

Would you agree that this value is exceptional, when it should usually be in [2. 6]? I conclude adding another slider for n does not make too much sense.

I find it more probable to assume a different value for the other unknown, H0, near 10.5. If your magazine provides H0, you can use it directly. Else just add a simple correction to absolute_magnitude.

#15 obrazell

Surely that assumes H is known accurately which it isn't. Comets also show different slope paratmeters pre and post perihelion so it is all pretty much guesswork to predict a magntitude. Taking into account observations also assume the observer is trustworthy as you may get wide scatter. Any comet magnitude prediction is only going to be guess work. If you get the position right then go and have a look.

#16 catalogman

Yes, that's the canonical formula that is also used in Stellarium. And thanks for the insight into the meaning of the slope parameter.

Now, e.g. see Dec. 5, 2018:
d=0.1AU. Log d=-1.

r=1.06AU. log r

0.025 which is pretty small. This second factor in this distance only has a small influence near r=1.
If we take MPC's H0=14 for real we have

H=14-5+2.5*0.025*n =9+0.0625n.

With a reported estimate of H=5.5 this reveals: (numerically don't believe it!)

-3.5=0.0625n, or n= - 56.

Would you agree that this value is exceptional, when it should usually be in [2. 6]? I conclude adding another slider for n does not make too much sense.

I find it more probable to assume a different value for the other unknown, H0, near 10.5. <snip>

Yes, from your values of H = 5.5, r = 1 A.U., and d = 0.1 A.U., we find that

H0 = 5.5 - 5 log (0.1) - 2.5 * 4 * log(1) = 10.5

Use another observation to find a more accurate value:

On Oct 30.41, H = 8.4, r = 1.2016 A.U., d = 0.2804 A.U. Toggling the power index gives the range

H0 = 8.4 - 5 log (0.2804) - 2.5 * 2 * log(1.2016) = 10.76
H0 = 8.4 - 5 log (0.2804) - 2.5 * 4 * log(1.2016) = 10.36
H0 = 8.4 - 5 log (0.2804) - 2.5 * 6 * log(1.2016) = 9.97

As a check, let's try to predict the magnitude at the time of this post: r = 1.1133 A.U., d = 0.1603 A.U., so

H = 10.76 + 5 log (0.1603) + 2.5 * 2 * log(1.1133) = 7.02
H = 10.36 + 5 log (0.1603) + 2.5 * 4 * log(1.1133) = 6.85
H = 9.97 + 5 log (0.1603) + 2.5 * 6 * log(1.1133) = 6.69

The current magnitude is actually 5.8.

so the power-law is not an accurate model of the nucleus (or the observation is not accurate). But the predicted magnitude, 10 weeks after the initial determination on Oct 30.41, is still much better than the mag = 10.7 value reported by CdC (and other programs that use the MPC). So, as already suggested above, the MPC's value of H0 is not accurate at this apparition. The observer should have the option to change both n and H0.

As a side note, here's more about the slope parameter for asteroids:

Notice that function H(alpha) for the reduced magnitude of an asteroid is the comet formula for n = 2 (reflection from a constant mass).


Calculating Apparent Magnitudes

You might be wondering what exactly the number you read off for an apparent magnitude tells you about the relative brightness of different objects. Magnitudes are built on what’s known as a logarithmic scale which allows us to compare objects with vastly different brightness without using incredibly large numbers. The magnitude scale works in such a way that an increase of 1 in magnitude corresponds to a decrease in brightness by a factor of about 2.5. In other words, an object with a magnitude of 5 is 2.5 times fainter than an object with a magnitude of 4.

The physical property that magnitude actually measures is flux, the amount of light that arrives in a given area on Earth in a given amount of time. Abbreviated as f , flux relates to SDSS magnitudes, m, in the following way:

M = 22.5 – 2.5 x log10 ( f)

The zero-point of this scale (the relative point to which other brightnesses are compared) is the flux of a standard source, which has a magnitude defined to be 22.5.

A variation of the equation above can be used to relate the difference between any two objects’ magnitudes (m1 and m2) to their flux ratio, f1/f2:

M1 – m2 = -2.5 x log10 (f1/f2)

The Sun, which is 14 units of apparent magnitude brighter than the full moon, is almost 400,000 times brighter if you compare the intensity of their light directly (this is probably not surprising, since we can safely look at the moon but not at the Sun). So now you can appreciate how magnitudes help us compare objects in the sky that have extremely different brightnesses without having to use enormous numbers.


Distance and Displacement problems

Problem 1:
A boat sailing through a river moved eastward for 5 km, then cross the river by moving 3 km southward. On reaching the other side it moved westward through 1 km and reached the jetty. Find the distance covered and displacement of the boat.

Solution:

From the fig, AB = 5 km
ED = BC = 3 km
CD = 1 km
∴ AE = 5 km – 1 km = 4 km

Distance covered = AB + BC + CD
= 5km + 3km + 1 km
= 9 km


So the distance covered by the boat is 9 km and the displacement is 5 km

Problem 2:
A car moving along in a straight highway from point P to point Q to point R and to point S, then back to point Q and finally to the point R as shown in the figure below.
a) Find the distance travelled by car.
b) Find the displacement of the car.

Ans:

Given the distances, PQ = 3 km, QR = 5 km and RS = 7 km
Also, SQ = 7 + 5 =12 km,
PR = 3 + 5 = 8 km

a) Distance travelled by the car = PQ + QR + RS + SQ + QR
= 3 + 5 + 7 + 12 + 5
= 32 km

b) Displacement of the car = the shortest distance between the final point R and the initial point P
= PR
= 8 km

Problem 3:
A person walks along the path of a rectangle from point P to point R as shown in the below figure.
a) Find the distance travelled by the person.
b) Find out the magnitude of the displacement of the person.

Ans:

Given the distances, PQ = 5 km,
QR = 2 km

a) The distance travelled by the person = PQ + QR
= 5 + 2
= 7 km

b) The magnitude of displacement is equal to the shortest distance between the final point R and the initial point P, which is equal to the diagonal PR and can be calculated using Pythagora’s theorem.
i.e., magnitude of displacement = PR
By Pythagora’s theorem, PR 2 = PQ 2 + QR 2

Problem 4:
A motorcycle rides from point P to Q to R to S and finally to P in a circular path as shown in the below figure.
Find a) the distance travelled by motorcycle.
b) the displacement.

Answer:

Given radius, r = 5 km

a) Here, the distance travelled by motor cycle = circumference of the circle (∵ the motor cycle moves one complete rotation)
= 2πr
= 2 × 3.14 × 5
= 31.4 km

b) Here, the initial point is P and the final point is also P. Therefore, there will be no change in position and hence the displacement is equal to zero.

Problem 5:
A vehicle moves from point P to Q to R to S in a circular path as shown in the below figure.
a) Find the distance travelled by the vehicle.
b) Find out the magnitude of the displacement of the vehicle.


Ans:

Given, radius, r = 8 km

a) Here, the vehicle moves only 3/4 th of one rotation.
∴, the distance travelled by the vehicle = 3/4 th of the circumference of the circle.
= (3/4)2πr
= (3/4) × 2 × 3.14 × 8
= 37.68 km

b) Here the initial point is P and the final point is S.
∴, The magnitude of displacement is equal to the shortest distance between the final point S and the initial point P, which is equal to the distance PS and can be calculated using Pythagoras theorem to the triangle POS as shown in the below figure.

Problem 6:
Consider an object moving in a straight line. The distances travelled by the object from the origin with respect to time are shown in the figure given below. The different parts of its motion are represented by P, Q, R, S, T, U, V and W. Find a) The distance travelled by the object in the first 2 seconds. b) the distance travelled by the object in the first 4 seconds. c) the distance travelled by the object in the first 6 seconds. d) the distance travelled by the object in the first 8 seconds. e) the distance travelled by the object in the first 9 seconds. f) total distance travelled by the object in 14 seconds and g) displacement in 14 seconds.

Ans:

From the distance time graph provided in the question, we can find out the distance and displacement for different time intervals. From the fig,
a) The distance travelled by the object in first 2 seconds = 60 m
b) The distance travelled by the object in first 4 seconds = 90 m
c) The distance travelled by the object in first 6 seconds = 90 m
d) The distance travelled by the object in first 8 seconds = 150 m

e) Distance travelled by the object in 9 seconds = 150 + (150 – 120 )
= 150 m + 30 m
= 180m.
That is, from t = 0 to t = 8 seconds, the object has moved a distance of 150 m from the origin and from t = 8 to t = 9 sec, the object has travelled back a distance of 30 meters. So the total distance will be 150 + 30 = 180 m.

f) The distance travelled by the object in 14 seconds = Total distance travelled by the object.
= 150 + 30 + 120
= 300 m.
That is, from t = 0 to t = 8 seconds, the object has moved a distance of 120 m from the origin and from t = 8 to t = 14 sec, the object has come back to its initial position and travelled a distance of 150 meters again. So total distance travelled = 150 m + 150 m = 300 metre.

g) Since the object has come back to its initial position, the total displacement is zero.

Problem 7:
The movement of objects P, Q, R, S and T is marked on a scale as shown in the below figure. Find the distance and displacement covered by each object.

Answer:

a) Consider the object P

Object P had an initial position of 1 metre and a final position of 4 metres.
∴ displacement of object P, Δxp= final position – initial position
= 4 – 1
= +3 metres.
Distance travelled by object P = total path covered covered by P
= 3 metres

b) Consider the object Q

Object Q had an initial position of 11 metres and a final position of 7 metres.
∴ displacement of object Q, Δxq= final position – initial position
= 7 – 11
= -4 metres.
Distance travelled by object Q = total path covered by Q
= 4 metres

c) Consider the object R

Object R had an initial position of 0 metres and a final position of 6 metres.
∴ displacement of object R, Δxr= final position – initial position
= 6 – 0
= +6 metres.
Distance travelled by object R = total path covered by R
= 6 + 3 + 3
= 12 metres

d) Consider the object S

Object S had an initial position of 8 metre and final position of 7 metres.
∴ displacement of object S, Δxs= final position – initial position
= 7 – 8
= –1 metre.
Distance travelled by object S = total path covered by S
= 3 + 4
= 7 metres

e) Consider the object T

Object T had an initial position of 7 metres and final position of 8 metres.
∴ displacement of object T, Δxt= final position – initial position
= 8 – 7
= 1 metre.
Distance travelled by object T = total path covered by T
= 4 + 3
= 7 metres.

Problem 8:
The horizontal position of a car in kilometres over time is shown below.

a) Find the displacement and the distance travelled by car between 1 hour and 3 hours.
b) What is the displacement and distance covered by the car between 3 hours and 5 hours?
c) Calculate the displacement and the distance travelled by car between 5 hours and 9 hours.
d) Find the displacement and the distance covered by the car between 3 hours and 9 hours.
e) Find the total displacement and the distance covered by the car.

Answer:

From the position-time graph given, we can calculate the distance and displacement for different time intervals.

a) Between 1 hour an 3 hour, the car had an initial position of 40 km and a final position of 40 km.
Displacement of the car between 1 hour and 3 hours, Δx = final position – initial position = 40 – 40 = 0 km .

Distance travelled by car between 1 hour and 3 hours = total path covered between 1 hour and 3 hours = 0 km ( Since that portion of the graph is a straight line parallel to the x-axis).

b) Similarly, the car had an initial position of 40 km and a final position of 160 km between 3 hours and 5 hours.
∴ Displacement of the car between 3 hour and 5 hours , Δx = final position – initial position = 160 – 40 = 120 km .

Distance covered by the car between 3 hour and 5 hours = total path covered between 3 hours and 5 hours = 120 km .

c) Similarly, the car starts at an initial position of 160 km and ends at a final position of 0 km during the period 5 hours and 9 hours.
∴ Displacement of the car between 5 hours and 9 hours , Δx = final position – initial position = 0 – 160 = -160 km ( displacement is negative means the car moves in the opposite or negative direction).

Distance covered by the car between 5 hours and 9 hours = total path covered between 5 hours and 9 hours = 160 km (distance is always positive)

d) Similarly, between the time 3 hours and 9 hours, the initial and final position of the car is 40 km and 0 km respectively.
Displacement of the car between 3 hours and 9 hours, Δx = final position – initial position = 0 – 40 = -40 km ( displacement is negative means the car moves in the opposite or negative direction).

From the graph, it is clear that the car travels in two segments between 3 hours and 9 hours. i.e., the car starts at 40 km at 3 hours and moves to 160 km at 5 hours, travelling a distance of 120 km. Also, the car starts at 120 km at 5 hours and moves to 0 km at 9 hours, travelling a distance of 160 km.
Distance covered by the car between 3 hours and 9 hours = total path covered between 5 hours and 9 hours = 120 km + 160 km = 280 km.

e) While considering the total motion, the car starts and ends at the same position of 0 km. That is, its initial position is 0 km at 0 hour and final position is also 0 km at 9 hours.
Total displacement of the car , Δx = Displacement of the car between 0 hours and 9 hours= final position – initial position = 0 – 0 = 0 km.

Also, while considering the total motion, it is clear that the car travels in four segments between 0 hours and 9 hours. i.e., the car starts at 0 km at 0 hours and moves to 40 km at 1 hour, travelling a distance of 40 km. Also, the car is stationary between the time 1 hour and 3 hours, thus the distance covered in this segment is 0 km. Again, the car starts at 40 km at 3 hours and moves to 160 km at 5 hours, travelling a distance of 120 km. Finally, the car starts at 120 km at 5 hours and moves to 0 km at 9 hours, travelling a distance of 160 km.
Total distance covered by the car = total path covered between 0 hours and 9 hours = 40 + 0 + 120 + 160 = 320 km

Problem 9:
A boy rides a bicycle back and forth along the ground and the horizontal position of the bicycle is given below.
a) Find the displacement and the distance covered by the bicycle between 0 second and 30 seconds.
b) Calculate the displacement and distance travelled by bicycle between 0 second and 40 seconds
c) Calculate the displacement and distance travelled by bicycle between 30 seconds and 50 seconds.
d) What is the total displacement and distance travelled by bicycle?

Answer:

The distance and displacement for different time intervals can be found out from the position vs time graph given.

a) The bicycle had an initial position of 60 metres and a final position of –30 metres between 0s and 30 seconds.
∴ Displacement of the bicycle between 0 second and 30 seconds , Δx = final position – initial position = -30 – 60 = -90 m.

Also, from the p-t graph given, it is clear that the bicycle travels in two segments between 0 second and 30 seconds. i.e., the bicycle starts at 60 m at 0 second and moves to – 30 metres at 15 seconds, travelling a distance of 90 m. Also, the bicycle is stationary between the time 15 seconds and 30 seconds, thus the distance travelled in this segment is 0 m.
∴ Distance covered by the bicycle between 0 second and 30 seconds = total path covered between 0 seconds and 30 seconds = 90 + 0 = 90 m.

b) Similarly, between the time 0 seconds and 40 seconds, the bicycle had an initial and final position of 60 metres and 30 metres respectively.
∴ Displacement of the bicycle between 0 second and 40 seconds , Δx = final position – initial position = 30 – 60 = -30 m.

Also, from the graph, it is clear that the bicycle travels in three segments between 0 second and 40 seconds. i.e., the bicycle starts at 60 m at 0 second and moves to – 30 metres at 15 seconds, travelling a distance of 90 m. Also, the bicycle is stationary between the time 15 seconds and 30 seconds, thus the distance travelled in this segment is 0 m. Again, the bicycle starts at –30 metres at 30 seconds and moves to 30 metres at 40 seconds, covering a distance of 60 metres.
∴ Distance covered by the bicycle between 0 second and 40 seconds = total path covered between 0 seconds and 40 seconds = 90 + 0 + 60 = 150 m.

c) Similarly, the bicycle starts at an initial position of -30 metres and ends at a final position of 0 metres during the period 30 seconds and 50 seconds.
∴ Displacement of the bicycle between 30 second and 50 seconds , Δx = final position – initial position = 0 – -30 = +30 m.

Also, from the graph, it is clear that the bicycle travels in two segments between 30 second and 50 seconds. i.e., the bicycle starts at -30 m at 30 seconds and moves to 30 metres position at 40 seconds, travelling a distance of 60 m. Also, the bicycle starts at 30 metres at 40 seconds and moves to 0 metres position at 50 seconds, covering a distance of 30 metres.
∴ Distance covered by the bicycle between 0 second and 50 seconds = total path covered between 0 seconds and 50 seconds = 60 + 30 = 90 m.

d) Now while considering the total motion, the bicycle starts at 60 metres and ends at 0 metres.
∴ Total displacement of the car , Δx = Displacement of the car between 0 second and 50 seconds = final position – initial position = 0 – 60 = -60 m.

Also, from the graph, it is clear that the bicycle travels in four segments between 0 second and 50 seconds. i.e., the bicycle starts at 60 m at 0 second and moves to – 30 metres at 15 seconds, travelling a distance of 90 m. Also, the bicycle is stationary between the time 15 seconds and 30 seconds, thus the distance travelled in this segment is 0 m. Again, the bicycle starts at –30 metres at 30 seconds and moves to 30 metres at 40 seconds, covering a distance of 60 metres. Finally, the bicycle starts at 30 metres at 40 seconds and moves to a 0 metre position at 50 seconds, covering a distance of 30 metres.
∴ Total distance covered by the bicycle = total path covered between 0 second and 50 seconds = 90 + 0 + 60 + 30 = 180 m.

Problem 10:
The position-time graph for an elevator travels up and down is given below. Find the distance and displacement of the elevator between 6 seconds and 21 seconds.

Answer:

The elevator had an initial position of -15 metres and a final position of 20 metres between 6 s and 21 seconds.
∴ Displacement of the elevator between 6 seconds and 21 seconds , Δx = final position – initial position = 20 – -15 = 35 m.

Also, from the graph, it is clear that the elevator moves in two segments between 6 second and 21 seconds. i.e., the elevator starts at -15 m at 6 second and moves to a 0-metre position at 9 seconds, covering a distance of 15 m. Again, the elevator starts at 0 m at 9 seconds and moves to 20 metres position at 21 seconds, covering a distance of 20 m.
∴ Distance covered by the elevator between 6 seconds and 21 seconds = total path covered between 6 seconds and 21 seconds = 15 + 20 = 35 m.

I hope the information in this article helps you to get a brief idea about distance and displacement. Also, I would love to hear your thoughts and feedback about this article via the comments section given below.

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