Register to reply 
Using kepler's laws to find mass of planet 
Share this thread: 
#1
May2205, 02:26 PM

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.



#2
May2205, 02:43 PM

Emeritus
Sci Advisor
PF Gold
P: 2,977

[tex]P^2=\frac{a^3}{M_p}[/tex] 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). 


#3
May2205, 02:47 PM

Sci Advisor
HW Helper
P: 3,033

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. 


#4
May2205, 03:04 PM

Sci Advisor
PF Gold
P: 1,542

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 rewrite the formula for circular velocity:
velocity = sqrt(Gravitational Constant * Mass / radius) into M = v^2 r / G G = gravitational constant = 6.673e11 compute your velocity from your radius and your period (in seconds). 


#5
May2205, 03:06 PM

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 2.24e6 as mass. with my previous equation is it right to AU and Years for A and P and M is KG?



#6
May2205, 03:21 PM

Sci Advisor
PF Gold
P: 1,542

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. 


#7
May2205, 03:21 PM

Emeritus
Sci Advisor
PF Gold
P: 2,977




#8
May2205, 03:26 PM

Emeritus
Sci Advisor
PF Gold
P: 2,977

Sounds like you might need to brush up on your unit conversion.



#9
May2205, 03:27 PM

P: 82

okay. i got it. i had all the units messed up. thanks guys :)



Register to reply 
Related Discussions  
Help with Kepler's Laws problem PLEASE!  Introductory Physics Homework  2  
Kepler's Laws  Introductory Physics Homework  4  
Satellite question. find mass of planet and weight on planet?  Introductory Physics Homework  2  
Kepler's Laws  Astronomy & Astrophysics  21  
Kepler's Laws  Astronomy & Astrophysics  28 