Question

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)

Answer 1

Answer:

5. 9GmM/(10R)

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 = $-\frac{GMm}{R}$

Therefore: W = Energy at earth surface - $\frac{GMm}{R}$

Energy at earth surface (E) at an altitude of 5R = $-\frac{GMm}{5r} +\frac{1}{2}mV^2$

But $V=\sqrt{\frac{GM}{5R} }$

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

W = E - PE

$W=-\frac{GMm}{10R}-(-\frac{GMm}{R})=-\frac{GMm}{10R}+\frac{GMm}{R}=\frac{9GMm}{10R} \\W=\frac{9GMm}{10R}$