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

Working from this: http://www.astro.umass.edu/~weinberg/a114/handouts/concept1.pdf \[ P^2=\frac{a^3}{M_s}\] The semi-major axis of the ellipse will be the average of R1 and R2 so \[ P^2=\frac{(\frac{R_1+R_2}{2})^3}{M_s}\ \ \ P=\sqrt{\frac{(\frac{R_1+R_2}{2})^3}{M_s}}\]

Answer 2

But you may need to use G to convert from R in terms of meters to a in terms of Astronomical Units or something.

Answer 3

Here is the formula which combines Kepler's 3rd law and Newton's law of gravitation:
G · m · t² = 4 · π² · r³
(where G=6.67x 10^-11, 'm' is in kilograms 't' is in seconds 'r' is in meters

Let's use some REAL numbers and say the planet is Mars orbits the Sun in 1.8808 years (or 59,354,000 seconds)
Its average distance from the Sun is 227,939,100,000 meters and the Sun's mass is 1.98855x10^30 kilograms.
We must solve for time and so
t = SquareRoot [4*PI²*r³ / G*m]
t = SqRoot [39.4784 * (2.27939x 10^11)³ / (6.67x 10^-11 * 1.989x10^30)]
t = Sq Root [(39.4784 * 1.18428 x 10^34) / 1.3267x10^20]
t = Sq Root [4.675x10^35 / 1.3267x10^20]
t = Sq Root [3.524 x 10^15]
t = 59,364,000 seconds

Source:
http://1728.org/kepler3a.htm
(formulas and calculator)