Calculating Work to Lift Satellite to Altitude

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Discussion Overview

The discussion revolves around calculating the work required to lift a 1000-kg satellite to an altitude of 2*10^6 m above the Earth's surface. Participants explore the application of gravitational force and work formulas, addressing both theoretical and mathematical aspects of the problem.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant poses the initial question about the work required, referencing the gravitational force formula and seeking clarification on the variables involved.
  • Another participant suggests that when force is variable, the work should be calculated using an integral, providing a specific integral setup for the problem.
  • There is a discussion about whether the masses involved (M and m) can be treated as constants and pulled out of the integral, with some participants agreeing on this point.
  • A participant expresses confusion regarding their calculated answer compared to the textbook answer, questioning the correctness of their approach.
  • Another participant points out a potential mathematical error in the calculations of the participant who disagrees with the textbook answer.
  • Clarification is provided regarding the upper limit of the integral, with a participant explaining that the total distance from the center of the Earth must include both the Earth's radius and the altitude of the satellite.
  • One participant concludes that the problem is resolved by adjusting the upper limit to the correct value.

Areas of Agreement / Disagreement

Participants generally agree on the need to adjust the upper limit of the integral to include the Earth's radius and the satellite's altitude. However, there remains disagreement regarding the correctness of individual calculations and the resulting answers, with some participants asserting they match the textbook while others do not.

Contextual Notes

Participants express uncertainty about specific mathematical steps and the interpretation of variables, particularly regarding the limits of integration and the treatment of constants.

-EquinoX-
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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
[tex]W= \int f(x)dx[/tex]

Here,
[tex]W= 6.67*10^{11} \int_{6.4*10^5}^{8.4*10^6}\frac{Mm}{r^2}dr[/tex]
 
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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