What is the change in gravitational potential energy

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Homework Statement


What is the change in gravitational potential energy of a 6200 kg satellite that lifts off from Earth's surface into a circular orbit of altitude 2500 km? What percent error is introduced by assuming a constant value of g and calculating the change in gravitational potential energy from m⋅g⋅Δh?


Homework Equations


[tex]E_{p}=mgh[/tex]
[tex]F_{g}=G((m_{1}m_{2})/(r^{2}))[/tex]


The Attempt at a Solution


Well, the change in potential energy is simply mgh = 6200kg*9.81m/s/s*2500000m = 152055000000 which is the change in p energy.. then the rest i am lost.. i know there would be an error because gravity is not 9.81 m/s as you go on, it gets less and less as you leave earth..

thanks for your time
 

Answers and Replies

  • #2
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You need to calculate the work done to get to the potential energy for the two models - work is the integral of the force needed along the path you take. The constant g integral is easy since the force is assumed constant, leading to the familiar mgh. Since the force in the second model depends on R you need to do the definite integral between the starting point and the end point...
 
  • #3
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You need to calculate the work done to get to the potential energy for the two models - work is the integral of the force needed along the path you take. The constant g integral is easy since the force is assumed constant, leading to the familiar mgh. Since the force in the second model depends on R you need to do the definite integral between the starting point and the end point...

We're not doing work yet, just momentum and energy.

Thanks
 
  • #4
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We're not doing work yet, just momentum and energy.

Thanks

work done for taking the satellite from R1 to R2 = difference in potential energy between these two locations.
 
  • #5
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work done for taking the satellite from R1 to R2 = difference in potential energy between these two locations.

yes, but for the first part i just use mgh? to find the change in p energy, mg(hf-hi), hi = 0

then for the percent error, i don't get what you mean...
 
  • #6
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yes, but for the first part i just use mgh?

right, because the integral from R1 to R2 over mg is simply mg*(R2-R1)
 
  • #7
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right, because the integral from R1 to R2 over mg is simply mg*(R2-R1)

okay thank you, therefore i get 152055000000 J, which equals 1.5*10^11 J

now what do i do for the second part? which equation would i use to find the actual value? cause i would then just find what the percent error is by dividing one by the other and subtracting that by 100 percent?

Thanks for your help!
 
Last edited:
  • #8
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The potential is the work done when taking the satellite from radius R1 to R2=R1+2500km. Putting them on a similar footing the work is
(1) [tex]\int_{R_1}^{R_2} dr G m_1 m_2 / R_1^2[/tex]
(2) [tex]\int_{R_1}^{R_2} dr G m_1 m_2 / r^2[/tex]
Where [tex]m_1[/tex] is the mass of earth, [tex]m_2[/tex] the mass of the satellite, [tex]G[/tex] the gravitational constant, [tex]R_1[/tex] an average radius of the earth.

Are you able to do these integrals?
 

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