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Classical Mechanics Level 5

Calculate the work that is done in Joules in launching a satellite of mass 1000 kg into an orbit above the Earth, at an elevation equal to the radius of the Earth?

Details and Assumptions

  • Earth’s mass = 6 × 1 0 24 6\times 10^{24} kg
  • Earth’s radius = 6.4 × 1 0 6 6.4 \times 10^6 m
  • You may assume that all the work applied to the satellite is transformed into potential energy.


The answer is 3.12E+10.

Mardokay Mosazghi by Mardokay Mosazghi , 22, USA

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2 solutions

Ameya Salankar
Jan 26, 2015

Instead of using G M m h R ( R + h ) \frac{GMmh}{R(R+h)}

we can simplify it a little and...

m g R h R + h \frac{mgRh}{R+h}

Put h = R h=R ,

m g R 2 2 R \frac{mgR^2}{2R}

which is

m g R 2 \frac{mgR}{2}

Substituting the parameters, we have the answer as 3.12 × 1 0 10 \boxed{3.12 \times 10^{10}} .

5 Helpful 0 Interesting 0 Brilliant 0 Confused

For some reason, I decided to do it by integration and somehow got it wrong. Serves me right, too. I used

W = ∫ R 2 R F G ( r ) d r = G M m 2 R W = \int_{R}^{2R} F_{G}(r) \ dr = \frac{GMm}{2R}

Somehow I got G M m 4 R \frac{GMm}{4R} at first.

Jake Lai - 6 years, 1 month ago

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i agree i used integration to solve this problem and it turned out fine

Mardokay Mosazghi - 5 years, 8 months ago
Pietro Pelliconi
Oct 21, 2014

Since the difference in work is equal to the difference in potential energy, W=GmM(1/r 1-1/r 2)

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Exactly how I did it. I wonder why this is worth 400 points?

Steven Zheng - 6 years, 5 months ago

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@Steven Zheng The problem was originally a different question, I had so many disputes and had to change it even though 5 people got it right.

Mardokay Mosazghi - 6 years, 4 months ago

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The question would be better if the third assumption was removed. You don't even have an orbit if you have no velocity. Work done to "launch into orbit" must provide energy for elevation to altitude PLUS energy for escape velocity - Kinetic + Potential. I think the answer should then be about 4.7E10 Joules - about 1.56E10 J for velocity and about 3.13E10 J for elevation. We are already ignoring the changing mass of the rocket as fuel is used up - make that the third assumption if there must be another detail.

Bob Kadylo - 5 years, 8 months ago

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