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Determine W (or fuel energy) required to launch a satellite of mass m at rest from a launching pad placed at the surface earth, to a circular orbit where the radius of the orbit is r = 5 R where R is the radius of the earth. Ignore the effect of the Earth’s rotation. Take the mass of the Earth to be M. The fuel energy required is:_______.

1. 5GmM/ (4R)
2. 3GmM/(2R)
3. 7GmM/(6R)
4. 2GmM/R
5. 9GmM/(10R)
6. 5GmM/(6R)
7. 11GmM/(10R)
8. 3GmM/(4R)

User Tarequeh
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1 Answer

4 votes

Answer:

5. 9GmM/(10R)

Step-by-step explanation:

m is the mass of the satellite

M is the mass of the earth

W is the energy required to launch the satellite

Energy at earth surface = Potential energy (PE) + W

W = Energy at earth surface - Potential energy (PE)

But PE =
-(GMm)/(R)

Therefore: W = Energy at earth surface -
(GMm)/(R)

Energy at earth surface (E) at an altitude of 5R =
-(GMm)/(5r) +(1)/(2)mV^2

But
V=\sqrt{(GM)/(5R) }

Therefore:
E=-(GMm)/(5R)+(1)/(2)m(\sqrt{(GM)/(5R) } )^2= -(GMm)/(5R)+(GMm)/(10R) = -(GMm)/(10R)

W = E - PE


W=-(GMm)/(10R)-(-(GMm)/(R))=-(GMm)/(10R)+(GMm)/(R)=(9GMm)/(10R) \\W=(9GMm)/(10R)

User Ohad Sharet
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