Examstyle Questions: Gravitational Potential Energy

This question is about gravitational field and potential near the Earth.
The gravitational potential V_{grav} due to the mass of an approximately spherical body is given by the expression.
V_{grav} = GM R where
M is the mass of the body
R is the distance from the centre of the body.
a) Show that the gravitational potential at the Earth's equator is about 6.25 x 10^{7} J kg^{1}.
mass of Earth = 5.98 x 10^{24} kg
radius of Earth at the Equator = 6.38 x 10^{6} m
radius of Earth at the poles = 6.36 x 10^{6} m
gravitational constant G = 6.67 x 10^{11} N m^{2} kg^{2}.
(2 Marks)
b) Show that the magnitude of the gravitational potential at one of the poles is about 1.003 times the magnitude at the equator.
(2 Marks)
c) Explain why gravitational potential is always negative.
(2 Marks)
(Marks available: 6 Marks)

a) The Moon's gravitational field at its surface id 1.70 Nkg^{1} and its mean radius is 1.74 x 10^{6} m. Calculate the mass of the Moon?
(3 Marks)
b) A rocket reaches a height of 100 m on the Earth. Ignoring the effect of air resistance on Earth, and assuming it operates in the same way as the Moon, caluculate the height that it would reach on the Moon.
Assume the Earth's gravitational field = 9.81 Nkg^{1} and the Gravitational constant (G) = 6.67 x 10^{11}m^{3} kg^{1} s^{2}.
(3 Marks)
(Marks available: 6)
Answer outline and marking scheme for question:

a) V_{grav}= 6.67 x 10^{11} x 5.98 x 10^{24}/ 6.38 x 10^{6} (1 Mark) = 6.252 x 10^{7} (1 Mark) J kg^{1}
Own value needed.
(2 Marks)
b) Calulating V_{grav} = 6.27 x 10^{7} (1 Mark) J kg^{1} using this to give ratio 6.27/6.25 = 1.003(2) (1 Mark)
Own value needed.
OR: explanation leading to 6.38/6.36 = 1.0031 (1 Mark)
(2 Marks)
c) Gravity is always attractive (AW) (1 Mark) hence it always takes energy/work to seperate gravitationally bound masses. At infinity the energy 'stored' is zero therefore an object in a field will be in a potential well. (1 Mark) (AW)
(2 Marks)
(Marks available: 6)

a)
mg_{m} = GmM (1 Mark) r^{2} So, M = gm.r^{2} = 1.70 x (1.74 x 10^{6})^{2} (1 Mark) G 6.67 x 10^{11} M = 7.72 x 10^{22} kg (1 Mark)
(3 Marks)
b) Potential Energy at max height is the same on the Earth and the Moon, so
Mgmhm = mg_{E} h_{E} (1 Mark)
(where M= mass of rocket, g is gravitational field strength and h is height reached on Moon (m) and Eart (E))
c)
So, hm = g_{E} h_{E} = 9.81 x 100 (1 Mark) Gm 1.70 Height on Moon = hm = 577 m (1 Mark)
(3 Marks)
(Marks available: 6)