Q4.11 Gravitational escape energy (S)
Calculate the amount of energy required to escape from the surface of (a) Earth’s moon and (b) Mars, relative to that required to escape from the surface of Earth.
[Data: M, M
, M
R, R
, R
]
Hint
Reveal
Solution
Reveal
Hide

![\[ U = -\frac{G M m}{r} \]](mastermathpng-6.png)
![\[ \frac{ E_{\rm moon} }{ E_{\rm earth} }= \frac{ G M_{\rm moon}m/r_{\rm moon} }{ G M_{\rm earth}m/r_{\rm earth} } =\frac{ M_{\rm moon} r_{\rm earth} }{ M_{\rm earth} r_{\rm moon} } =\frac{ (7.36\times 10^{22} kg)\times(6.37\times 10^{6} m) }{ (5.98\times 10^{24}kg)\times (1.74\times 10^{6}m) }=0.0451 \]](mastermathpng-7.png)

Solution
From Key Point 3.14 we have that the gravitational potential energy of a particle of mass m a distance r from a planet of mass M is
![\[ U = -\frac{G M m}{r} \]](mastermathpng-6.png)
For the mass to ‘escape’ from the planet it must attain a separation r=∞ from the planet so that U=0. Thus
(a)
![\[ \frac{ E_{\rm moon} }{ E_{\rm earth} }= \frac{ G M_{\rm moon}m/r_{\rm moon} }{ G M_{\rm earth}m/r_{\rm earth} } =\frac{ M_{\rm moon} r_{\rm earth} }{ M_{\rm earth} r_{\rm moon} } =\frac{ (7.36\times 10^{22} kg)\times(6.37\times 10^{6} m) }{ (5.98\times 10^{24}kg)\times (1.74\times 10^{6}m) }=0.0451 \]](mastermathpng-7.png)
(b)
![\[ \frac{ E_{\rm mars} }{ E_{\rm earth} }=\frac{ GM_{\rm mars} m / r_{\rm mars} }{ GM_{\rm earth}m/r_{\rm earth} } =\frac{ M_{\rm mars}r_{\rm earth} }{ M_{\rm earth}r_{\rm mars} } =\frac{ (6.54\times 10^{23}kg)\times(6.37\times 10^6 m) }{ (5.98\times 10^{24}kg)\times (3.397\times 10^{6}m) }=0.205 \]](mastermathpng-8.png)