Gravity on the height of 12740 km ?

[Amar]

Homework Statement

So the problem I have is calculating the gravity acceleration on the height of 12740km or 2 times the radial distance of Earth. The problem is relatively simple and I think that I have it right but the result doesn't match with the book.

$g_0=9,81 \frac {m}{s^2}$
$R=6370km$
$h=R=6370km$

Homework Equations

$g = g_0 \frac{R^2}{(R+h)^2}$

The Attempt at a Solution

Well given that I have everything except $g$ I could have just put everything in and be done with it. I went with canceling out the symbols and got $g = \frac {g_0}{2}$ This resulted in 4.905. When calculating everything the result was 2.4525. The result specified in the book is 1,09 m/s^2. I'm really confused :)

 

[gneill]

A height is a distance above the Earth's surface. You're given a height of 2R, so the radial distance is 3R...

 

[Amar]

A height is a distance above the Earth's surface. You're given a height of 2R, so the radial distance is 3R...

I must have explained it badly. The height given is 1R

EDIT: The title is wrong, sorry..

 

[gneill]

I must have explained it badly. The height given is 1R
.

That would make the book's answer incorrect...

 

[Amar]

That would make the book's answer incorrect...

But what about the double answer I got ? One is 4.905 the other 2.4525 ? Kinda weird that the the other is 2x smaller than the first :P

 

[gneill]

But what about the double answer I got ? One is 4.905 the other 2.4525 ? Kinda weird that the the other is 2x smaller than the first :P

Can you show your algebraic steps for your attempt by cancellation?

 

[Amar]

But what about the double answer I got ? One is 4.905 the other 2.4525 ? Kinda weird that the the other is 2x smaller than the first :P

Never mind, after another try at the calculation I got that 2.4525 is correct, thanks for the help :)