Calculating Work to Lift Satellite to Altitude

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SUMMARY

The discussion focuses on calculating the work required to lift a 1000-kg satellite to an altitude of 2 x 10^6 m above the Earth's surface. The gravitational force is defined by the formula F = GMm/r^2, where G is the gravitational constant (6.67 x 10^-11), M is the Earth's mass (6 x 10^24 kg), and m is the satellite's mass. The correct calculation involves integrating the gravitational force over the distance from the Earth's center, leading to the conclusion that the upper limit for the integral should be 8.4 x 10^6 m, which resolves discrepancies in the calculated work.

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How much work is required to lift a 1000-kg satelitte to an altitude of 2*10^6 m above the surface of the earth? The gravitational force is F = GMm/r^2, where M is the mass of the earth, m is the mass of the satellite, and r is the distance between them. The radius of the Earth is 6.4*10^6m, its mass is 6*10^24kg, and in these units the gravitational constant, G, is 6.67*10^-11.

I know that the formula of work is W = F * d

d is distance. What is r here? The distance between the satellite and earth? Is it the 2*10^6 and what is d here? Thanks
 
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Work= F*d if F is a constant. When F is a variable, you need
W= \int f(x)dx

Here,
W= 6.67*10^{11} \int_{6.4*10^5}^{8.4*10^6}\frac{Mm}{r^2}dr
 
M and m are also constant right? so we can pull that both outside from the integral??
 
Last edited:
Yes, the masses are constant.
 
why did I get the answer key from the textbook:

I got GMm * ((-1/8*10^6) + (1/6.4*10^6))

is that right??
 
-EquinoX- said:
why did I get the answer key from the textbook:

I got GMm * ((-1/8*10^6) + (1/6.4*10^6))

is that right??

Do you mean this is the answer in the book? Or this is the answer you got and it doesn't match the book?
 
this is the answer I got and it didn't matches the book
 
What's the answer in the book? How much are you off by?
 
the answer in the book is 1.489*10^10 and I got 1.042*10^10
 
  • #10
I think you might be making a math error. I got the same answer as the book.
 
  • #11
One thing I see is that in your answer you have the height of the satellite as 8x10^6, when it should be 8.4x10^6. I don't think it is really going to change it too much though.
 
  • #12
where did you get 8.4 from?? there isn't 8.4 in the question
 
  • #13
In your question, you have 6.4x10^6 m as the radius of the Earth. Add the 2x10^6 m altitude of the satellite to that, and you get 8.4x10^6 m as your upper limit.

You have to take all distances from the centre of the Earth.
 
Last edited:
  • #14
okay the problem is solved if I change it to 8.4*10^6
 

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