Planetary Motion with satellite

[teme92]

Homework Statement

A 20 kg satellite has a circular orbit with a period 2.4 h and radius 8.0×10⁶m around a planet of unknown mass. If the magnitude of the gravitational
acceleration on the surface of the planet is 8.0 m/s², what is the radius of the
planet?

Homework Equations

F = Gm₁m₂/r²
F_c = m₁v²/r
v=D/T, D=\pir

The Attempt at a Solution

So I got the velocity of the the satellite by using speed=distance/time where the distance is the circumference of the orbit or \pir and T is the period in seconds. I'm having trouble because there is an unknown mass of the planet and its radius. How do I go about finding this values? Any help is much appreciated.

 

[patrickmoloney]

This is from the 2011 paper, it's actually a lot easier than it looks.

T = period $T = (2.4)(3600) = 8640 s$

Okay so I did this question using Kepler's second law (law of periods). It's a very useful law and is worth learning.

<content deleted -- complete worked solutions are not permitted> -- gneill, PF Mentor

 

[teme92]

I thought the radius value was R_Total = R_Satellite + R_Planet.

Also using your method I got r = 5.3x10⁶. Did you use g=9.8 or is there some other difference between or values? Also thanks for the help.

 

[patrickmoloney]

It's not the acceleration of the Earth its the acceleration of the planet $a_g = 8.0 m/s^2$ So let's break this down, we use Kepler's second law, using a little bit of algebra this relation provides is with an approximation of the mass of the planet. Having the mass of the planet and the acceleration of the planet we have the radius of the planet.

$∑F = ma_g = \frac{GMm}{r^2}$ which implies $a_g = \frac{GM}{r^2}$

 

[Chestermiller]

What is the radial acceleration of the satellite at 8x10⁶ m? You know the radial acceleration at this location, you know radial acceleration at the surface, and you know that the radial acceleration is inversely proportional to the square of the radius.

Chet

 

[dauto]

This is from the 2011 paper, it's actually a lot easier than it looks.

T = period $T = (2.4)(3600) = 8640 s$

Okay so I did this question using Kepler's second law (law of periods). It's a very useful law and is worth learning.

<worked solution content deleted -- gneill, PF Mentor>

You're not supposed to solve the problem. Just explain how the solution works.

 

[teme92]

Ok thanks for clearing that up and all the help guys