Question

A planet moves in an elliptical orbit around the sun. The mass of the sun is Ms. The minimum and maximum distances of the planet from the sun are R1 and R2, respectively.
Using Kepler's 3rd law and Newton's law of universal gravitation, find the period of revolution P of the planet as it moves around the sun. Assume that the mass of the planet is much smaller than the mass of the sun.
Use G for the gravitational constant.

Answer 1

Even though im not giving you the whole answer im going to give you such a help that i know you will be able to fix this.Ok so let me state that  Kepler's third law states the square of the period is proportional to the cube of the average distance. So for that the formula you need is: T = sqrt { 4 * PI^2 /(GM) * [(R1 + R2)/2]^3 } 
I knoe this is going to help so that you can get through. 

Answer 2

Answer:

$2\pi \sqrt{\frac{a^{3} }{GMs } }$

Explanation:

Kepler's 3rd Law states that the square of the revolutionary period of a planet around the sun is proportional to the cube of the semimajor axis's orbit.

First we find the semimajor axis a. The semimajor axis of the ellipse is half of its main axis. The sun is in the focus of the elliptical orbit and the focus is on the main axis

a = $\frac{R1 + R2}{2}$

Then we find the period P. The period P is (2πr) / v, where r is the radius of the orbit and v is the velocity of the object. The distance traveled in one orbit divided by speed.

Then find the velocity of an object in the orbit of the radius by adjusting the magnitude of the centripetal acceleration equal to the magnitude of the acceleration due to gravity.

$a_{cent} = v^{2} /r$

v = $\sqrt{\frac{MsG}{r} }$

For a circle, the semi­major axis is just the radius.

$2\pi \sqrt{\frac{a^{3} }{GMs } }$

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Class : Senior High

Subject : Biology

Keywords

- Kepler's 3rd

- Period of revolution