Centripetal acceleration of a satellite orbiting Earth

[ride4life]

Homework Statement

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.

Homework Equations

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

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 :(

 

[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!

What does that give you? ​

 

[ride4life]

Ummm...
F_c=F_g
ma_c=F_g
ma_c=mg
a_c=g
?

 

[tiny-tim]

F=mg
g=GM/r²

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

 

[ride4life]

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

 

[ride4life]

ahhh...

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

 

[Matterwave]

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

 

[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 … ?

 

[ride4life]

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

 

[HallsofIvy]

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.

 

[ride4life]

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