# Centripetal acceleration of a satellite orbiting Earth

1. Apr 15, 2010

### ride4life

1. The problem statement, all variables and given/known data
A satellite is orbiting the Earth in a stable orbit of which the radius is twice that of the Earth. Find the ratio of the satellites cepripetal acceleration to g on the Earth's surface.

2. Relevant equations
centripetal acceleration = (v^2)/r or (4rpi^2)/T^2

3. The attempt at a solution
Well the only thing i can figure is the rafius of the satellite which is 12.8x10^6
Otherwise I have no idea where to start :(

2. Apr 15, 2010

### tiny-tim

Hi ride4life!

(try using the X2 tag just above the Reply box )

As usual, it's good ol' Newtons' second law that's needed, this time together with another of Newton's laws! :tongue2:

What does that give you?

3. Apr 15, 2010

### ride4life

Re: Gravitation

Ummm...
Fc=Fg
mac=Fg
mac=mg
ac=g
?????

Last edited: Apr 15, 2010
4. Apr 15, 2010

### tiny-tim

ok … now combine that with your original g = v2/r

5. Apr 15, 2010

### ride4life

Re: Gravitation

v2=GM/r
v2=(6.67x10-11x6x1024)/6.4x106
v=7907.67m/s

6. Apr 15, 2010

### ride4life

Re: Gravitation

ahhh...

ac=GM/r2
ac=(6.67x10-11x6x1024)/(2x6.4x106)2
ac=2.44

7. Apr 15, 2010

### Matterwave

Re: Gravitation

Remember to always include the units or else the answer makes no physical sense.

8. Apr 15, 2010

### tiny-tim

Sorry, not following this …

you're making it very complicated

The question only asks for the centripetal acceleration (as a multiple of g on the Earths' surface) …

you have centripetal acceleration = GM/r2, so … ?

9. Apr 15, 2010

### ride4life

Re: Gravitation

centripetal acceleration = 2.44
ratio compared to Earth's g which is 9.8 is 2.44/9.8 which is about 1:4

10. Apr 15, 2010

### HallsofIvy

Staff Emeritus
Re: Gravitation

It is only "about" 1:4 because of the over-complicated way you did that. That was tiny-tim's point. It isn't necessary to find the actual accelerations.

$$a= \frac{GM}{r^2}$$
If r= 2R (R is the radius of the earth) then
$$a= \frac{GM}{(2R)^2}= \frac{GM}{4r^2}= \frac{1}{4}\frac{GM}{r^2}= \frac{1}{4}g$$.

The centripetal acceleration of a satellite at radius twice the radius of the earth is exactly 1/4 g.

11. Apr 15, 2010

### ride4life

Re: Gravitation

ahhh...
i see what you mean, way less complicated than my way