Answer to Question #193522 in Field Theory for Freya

Question

 Q4- The Moon orbits the Earth along a path of radius 3.84 x 108 m. The mass of the Moon is 7.35 x 1022 kg, and the mass of the Earth is 6.00 x 1024 kg. (AC 1.2) a) The gravitational force of attraction between the Earth and the Moon keeps the Moon in orbit. Calculate the size of this force. b) A rocket of mass 42000kg is fired from Earth to the Moon. What is the net gravitational force on the rocket when it is 3 x 108 m from the centre of the Earth?

Answer 1

Answer

(A)

Gravitational force can be given as

$F=\frac{Gm_1m_2}{r^2}$

$F=\frac{(6.67\times10^{-11})(7.35\times10^{22})(6.0\times10^{24})}{(3.84\times10^8)^2}$

$F=2\times10^{20}N$

(B)

Force on rocket due to earth

$F_1=\frac{(6.67\times10^{-11})(42000)(6.0\times10^{24})}{(3\times10^8)^2}=186.79N$

Force on rocket due to moon

$F_2=\frac{(6.67\times10^{-11})(42000)(7.35\times10^{22})}{(0.84\times10^8)^2}=24.51N$

Net force

$F_{net}=186.79-24.51=162.28N$