# Using kepler's laws to find mass of planet

by jhson114
Tags: kepler, laws, mass, planet
 P: 82 There's a planet with a moon that orbits the plant in 30 days and moves at a distance of 370million meters from the center of the planet. i need to find the mass of the planet and i've used the following equation that was derived from the keplers laws: P = 2pi(a^(3/2))/(GM)^(1/2). P is the time it takes to or it and A is the average distnace from the planet. I've tried using P in days and years, and a in meters and AU. But i cant seem to arrive at a correct answer. maybe my equation is wrong? thanks in advance.
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PF Gold
P: 2,977
 Quote by jhson114 There's a planet with a moon that orbits the plant in 30 days and moves at a distance of 370million meters from the center of the planet. i need to find the mass of the planet and i've used the following equation that was derived from the keplers laws: P = 2pi(a^(3/2))/(GM)^(1/2). P is the time it takes to or it and A is the average distnace from the planet. I've tried using P in days and years, and a in meters and AU. But i cant seem to arrive at a correct answer. maybe my equation is wrong? thanks in advance.
If P is in years, M is in solar masses, and a is in AU, it's simply:

$$P^2=\frac{a^3}{M_p}$$

For your version of the equation, you should use some other consistent set of units (like meters, kilograms, and seconds; or centimeters, grams, and seconds).
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P: 3,031
 Quote by jhson114 There's a planet with a moon that orbits the plant in 30 days and moves at a distance of 370million meters from the center of the planet. i need to find the mass of the planet and i've used the following equation that was derived from the keplers laws: P = 2pi(a^(3/2))/(GM)^(1/2). P is the time it takes to or it and A is the average distnace from the planet. I've tried using P in days and years, and a in meters and AU. But i cant seem to arrive at a correct answer. maybe my equation is wrong? thanks in advance.
Your equation looks right. Do you know the mass of the moon? The mass in this equation is really the reduced mass of the planet moon combination.

If the orbit is elliptical, a is the semimajor axis of the ellipse. Since you are only given one distance from the planet, I assume you are supposed to be treating the obit as circular with a = 370million meters.

Edt
If reduced mass is relevant, then the distance would be to the center of mass, not to the other object, so this was not a helpful idea. I see you got it worked out now with a simpler calculation.

 Sci Advisor PF Gold P: 1,544 Using kepler's laws to find mass of planet If you can assume the orbit is circular, and the planet's mass >> moon's mass, you could also re-write the formula for circular velocity: velocity = sqrt(Gravitational Constant * Mass / radius) into M = v^2 r / G G = gravitational constant = 6.673e-11 compute your velocity from your radius and your period (in seconds).
 P: 82 i dont know the mass of the moon and we are to assume that its a circular orbit. the equation P^2 = a^3/M doesnt seem right because i get a 2.24e-6 as mass. with my previous equation is it right to AU and Years for A and P and M is KG?
 Sci Advisor PF Gold P: 1,544 You got the right answer except for one thing. That way gives solar masses and not kilograms. The Sun is 1.98911e30 kg. The method I gave you wants meters, seconds, and gives kilograms. Try it both ways to see if you get approximately the same answer. You should.
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