# Using Kepler's 3rd Law to find Period of Venus

• Rajveer97
In summary: You can use the two equations to find ##T_{Venus}## as$$T_{Venus}=\frac{2\pi}{\sqrt{Gm}}r_{Venus}^{3/2}$$
Rajveer97

## Homework Statement

Deduce, from the equations employed in Q4 and Q5, the exponent n in the equation: T = k rn where k is a constant and T is the period of a satellite which orbits at a radius r from a massive object in space. Hence, how long is the “year” on Venus if its distance from the Sun is 72.4% of the Earth’s?

## Homework Equations

The equations I used in the previous equations were simply a = (4(pi^2)r)/T^2 and a = GM/r^2 and basically combinations of the two

## The Attempt at a Solution

So I started this question by trying to find the distance of the Earth from the Sun using T^2 = (4pi^2)/Gm (m being mass of the Sun) to which I got the answer 2.75x10^11 km

Then I substituted the r in the Kepler's equation with 0.724Re (to find distance of Venus from the Sun) but my answers seemed to have all been wrong so either I'm making some silly arithmetical error or my entire approach is wrong. Any help and hints would be appreciated, thank you :)

Rejverr97:

The question asks you to write an equation of the form ##T=k r^n##, using the equations you had previously. That is, you need to combine those equations algebraically to express the period ##T## in terms of the distance ##r##. In the process, you'll find the exponent ##n## as well as the constant ##k## in terms of the constants in your previous equations.

Once you have that expression, you can work with ratios of the ##T## and ##r## values. If you do that, you should not need to work out intermediate values like the actual distance of Earth from the Sun in metres.

Does that help?

James R said:
Rejverr97:

The question asks you to write an equation of the form ##T=k r^n##, using the equations you had previously. That is, you need to combine those equations algebraically to express the period ##T## in terms of the distance ##r##. In the process, you'll find the exponent ##n## as well as the constant ##k## in terms of the constants in your previous equations.

Once you have that expression, you can work with ratios of the ##T## and ##r## values. If you do that, you should not need to work out intermediate values like the actual distance of Earth from the Sun in metres.

Does that help?

I should have mentioned, I did do that first step which gave me T = (2pi/root Gm) x r^3/2 . So the value I got for n was 3/2. Could you elaborate a bit more on what you mean later by working out the ratios of the T and r values? Because currently I can't think in my head how to do this without needing the distance of Earth from the Sun.

Rajveer97 said:
I should have mentioned, I did do that first step which gave me T = (2pi/root Gm) x r^3/2 . So the value I got for n was 3/2. Could you elaborate a bit more on what you mean later by working out the ratios of the T and r values? Because currently I can't think in my head how to do this without needing the distance of Earth from the Sun.
You don't need the absolute distance of either planet from the Sun. You are given the ratio of the two radii, and you are asked for Venus' "year" as measured in Earth years, i.e. you are asked for the ratio of the two years.
Combine that with the formula you obtained in the first part of the question, T=kr3/2.

Rajveer97 said:
I should have mentioned, I did do that first step which gave me T = (2pi/root Gm) x r^3/2 . So the value I got for n was 3/2. Could you elaborate a bit more on what you mean later by working out the ratios of the T and r values? Because currently I can't think in my head how to do this without needing the distance of Earth from the Sun.
Ok. So, you have
$$T_{Earth}=\frac{2\pi}{\sqrt{Gm}}r_{Earth}^{3/2},~~T_{Venus}=\frac{2\pi}{\sqrt{Gm}}r_{Venus}^{3/2}$$
Two equations. You know ##r_{Venus}/r_{Earth}=0.724## and ##T_{Earth}=1## year. Can you use the two equations to find ##T_{Venus}##?

## What is Kepler's 3rd Law and how does it help us find the period of Venus?

Kepler's 3rd Law, also known as the Law of Harmonies, states that the square of a planet's orbital period is proportional to the cube of its semi-major axis. This means that for any planet, the ratio of its orbital period (P) to the cube of its semi-major axis (a) will be the same. By using this law, we can calculate the period of Venus by knowing its semi-major axis.

## What is the semi-major axis of Venus?

The semi-major axis of Venus is 108,208,930 kilometers, or 0.723 astronomical units (AU). This is the average distance from Venus to the Sun.

## How accurate is using Kepler's 3rd Law to find the period of Venus?

Using Kepler's 3rd Law to find the period of Venus is quite accurate. In fact, it has been used successfully by scientists for centuries to predict the orbital periods of planets in our solar system with a high level of accuracy.

## Why is it important to know the period of Venus?

Knowing the period of Venus is important for understanding the dynamics of our solar system and how different planets interact with each other. It also helps us to make accurate predictions about future astronomical events, such as planetary alignments and transits.

## Can Kepler's 3rd Law be applied to other planets?

Yes, Kepler's 3rd Law can be applied to any planet in our solar system, as long as its orbital period and semi-major axis are known. It can also be used to calculate the orbital periods of other celestial bodies, such as moons and asteroids, as long as their semi-major axes are known.

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