Gravity on the height of 12740 km ?

  • Thread starter Amar
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  • #1
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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 :)
 

Answers and Replies

  • #2
gneill
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A height is a distance above the Earth's surface. You're given a height of 2R, so the radial distance is 3R...
 
  • #3
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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..
 
  • #4
gneill
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I must have explained it badly. The height given is 1R
.
That would make the book's answer incorrect...
 
  • #5
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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
 
  • #6
gneill
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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?
 
  • #7
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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 :)
 

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