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

I don't know how to approach this problem....

Answer 2

Using Kepler's 3rd law and Newton's law of universal gravitation:
\(\Large \frac{T^2}{R^3}=\frac{4 \pi ^2 }{GM_{sun}}\) as you can see, the mass of the planet doesn't matter.
solving for the period, T:
\(\Large T= \sqrt{\frac{4 \pi ^2 R^3 }{GM_{sun}}}\), where R is the distance between the sun and the planet, \(M_sun\) is the mass of the sun.

Answer 3

i am not sure why you would bother to invoke Kepler, that seems pointless to me. the way to do this is Newton's way , ie by by equating (1) GMm/r^2 with (2) mv^2/r, ie gravitation attraction between the 2 bodies = mass * angular acceleration of the smaller object.

en route, one, as Newton did, can show that T^2/R^3 is what will always happen provided one accepts Newton's inverse square law; but that is neither here not there in the context of this problem.

so, given formulae (1) & (2) above, and given that T (the period) = 2πR/v, you can then hack away at them and get to the answer provided above by the prev poster.

[the one caveat is that the math is really simple if you assume a circular orbit, and they are usually very nearly circular, because you can use mv^2/r for tots circular motion. but I noted somewhere that the period is not affected by the eccentricity of the elliptical orbit. so you just need to be sure to use the semi major axis as 'R" in the equation and that makes R in this case (R1 + R2)/2, the simple average.]

just my two. hope it helps in some way.