What is Kepler's Second Law and How Does it Relate to a Planet's Time Period?

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SUMMARY

This discussion focuses on the application of Kepler's Second Law to calculate the time period of a planet's orbit. The formula used, T^2 = 4*(pi)^2*(semi-major axis)^3 / GM, was misapplied by treating M as the mass of the planet instead of the central body. The correct interpretation reveals that the mass of the orbiting body does not affect the orbital period, which is solely determined by the semi-major axis. The calculated time period should be approximately 1.7 - 1.8 hours, not the erroneous 6.757*10^6 hours initially derived.

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  • Understanding of Kepler's Laws, particularly Kepler's Second Law
  • Familiarity with the concept of orbital angular momentum
  • Knowledge of elliptical orbits and their geometric properties
  • Basic proficiency in algebra and physics equations related to motion
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  • Learn how to calculate the area of an ellipse using semi-major and semi-minor axes
  • Explore the implications of Kepler's Third Law in different orbital scenarios
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Students studying physics, particularly those preparing for competitive exams like the IIT, as well as educators teaching orbital mechanics and celestial dynamics.

tanaygupta2000
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Homework Statement
A planet of mass 1 kg is moving under a central force in an elliptic orbit with semi major axis 1000
m and semi minor axis 100 m. The orbital angular momentum of the planet is 100 kg m^2 /s. Find the time period of the motion of the planet.
Relevant Equations
T^2 = 4*(pi)^2 *(semi-major axis)^3 / GM
By using the above formula I am getting value of time-period T as 6.757*10^6 hrs.
But the answer given is 1.7 - 1.8 hrs.
 
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Can you show your calculation?

What value of M did you use?

i think M represents the mass of the body being orbited. You have a 1kg planet orbited a larger body of mass M.
 
tanaygupta2000 said:
Homework Statement:: A planet of mass 1 kg is moving under a central force in an elliptic orbit with semi major axis 1000
m and semi minor axis 100 m. The orbital angular momentum of the planet is 100 kg m^2 /s. Find the time period of the motion of the planet.
Homework Equations:: T^2 = 4*(pi)^2 *(semi-major axis)^3 / GM

By using the above formula I am getting value of time-period T as 6.757*10^6 hrs.
But the answer given is 1.7 - 1.8 hrs.
How can a planet have a mass of only ##1kg##?
 
PeroK said:
How can a planet have a mass of only ##1kg##?
Or a semi-major axis of 1000m, semi-minor axis of 100m?

I'd say: to simplify the math for a homework assignment.
 
DaveC426913 said:
I'd say: to simplify the math for a homework assignment.

The maths is the same, whatever the numbers.
 
Only this much is provided in the question. I have no idea how to get solution.
 
tanaygupta2000 said:
Only this much is provided in the question. I have no idea how to get solution.
How is it that you've been given a problem that you have no idea how to solve. Did you miss class?
 
One piece of information you have that you haven't used yet is the angular momentum. Since you can't solve it on the basis of semimajor and semiminor axes, perhaps there's a reason there was other information given.

If you review your class notes, is it possible you may have been given a relationship between angular momentum and other quantities that might be useful here?
 
Annotation 2019-12-29 195605.png
 
  • #10
This is one of the IIT questions I was practicing.
 
  • #11
tanaygupta2000 said:
This is one of the IIT questions I was practicing.
So, it sounds like you want to be able to solve the problem without having studied the material, yes?
 
  • #12
phinds said:
So, it sounds like you want to be able to solve the problem without having studied the material, yes?
I thought this question can be solved using Kepler's laws, which I have studied.
 
  • #13
tanaygupta2000 said:
I thought this question can be solved using Kepler's laws, which I have studied.

One lesson you should learn here about Kepler's Laws is what M means in that equation. Apparently you treated it as the mass of the planet, but it is the mass of the central body. Which means that the mass of the orbiting body does not appear in the equation. All orbiting bodies of the same semi-major axis will have the same period (under the approximation that the central body has very large mass compared to the orbiting body).

This is very important.

Meanwhile, for solving this problem that means that you do not have sufficient information to apply Kepler's Third. So you need to find another equation that applies. And you also were given, as I said above, some additional information besides the orbital radius. Is it possible that additional information might be useful and another equation might be useful? What should you do when you find the initial equation you try is not useful?
 
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  • #14
There is a nifty way to approach this problem via Kepler's 2nd law (equal areas in equal times).

If you know the area contained within the elliptical orbit, which you should be able to find given that you are given both the semi-major and semi minor axes, then the period of the planet must be related to this area and the specific angular momentum (m2/s).
 
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  • #15
PeroK said:
The maths is the same, whatever the numbers.
I know, but it's still the most likely explanation.
 
  • #16
DaveC426913 said:
I know, but it's still the most likely explanation.

Although, is the mass of the planet relevant?

... I guess it is if you need to use the angular momentum!
 
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  • #17
tanaygupta2000 said:
I thought this question can be solved using Kepler's laws, which I have studied.
Follow @gneill's hint (Post #14). What does Kepler+s second law say?
 

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