How Is Work Calculated for a Falling Space Vehicle in Gravity?

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SUMMARY

The discussion focuses on calculating the work done by a falling space vehicle, specifically a 2500 kg vehicle descending from a height of 3100 km. The gravitational force is derived using Newton's law of universal gravitation, represented by the equation F_g = (m_1m_2)/(r^2)G. The work done is calculated through the integral W = ∫ F_g dl, leading to a final result of W = 5.1 x 10^10 Joules. The calculations and integral setup presented are confirmed to be correct by other forum participants.

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Homework Statement


A 2500 kg space vehicle, initially at rest, falls vertically from a height of 3100 km above the Earth's surface.


Homework Equations


[tex]\frac{m_1m_2}{r^2}G=F_g[/tex]
[tex]W=\int_a^b Fcos\theta dl[/tex]


The Attempt at a Solution


[tex]W=\int_a^b F_g dl[/tex]
[tex]r=r_{Earth} + r_{ship}[/tex]
[tex]W=\int_a^b \frac{m_1m_2}{r^2}G[/tex]
[tex]W=m_1m_2G\int_a^b r^{-2} dl[/tex]
[tex]W=m_1m_2G\bigg[-\tfrac{1}{r}\Big|_a^b\bigg][/tex]
[tex]W=m_1m_2G\bigg[-\frac{1}{r_{Earth} + r_{ship}}\Big|_a^b\bigg][/tex]
[tex]W=m_1m_2G\bigg[-\frac{1}{r_{Earth} + a}+\frac{1}{r_{Earth} + b}\bigg][/tex]
[tex]a=3.1*10^6m; b=0[/tex]
[tex]W=(2500)(5.98*10^{24})(6.67*10^{-11})\bigg[-\frac{1}{(6.38*10^6) + (3.1*10^6)}+\frac{1}{(6.38*10^6) + 0}\bigg][/tex]
[tex]W=5.1*10^{10}[/tex]

I believe this work is correct, can someone verify that my integral is correct please?
 
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