June'94 Paper 2 Question 2
(a) Define the term gravitational field strength
(b) State the numerical value and the unit of the gravitational field strength of the Earth at its surface.
(c) Why is it incorrect to call g(=9.8 ms-2) gravity?
(d) This part of the question is about the rotation of the Moon in a circular orbit around the Earth. You will need to use the following astronomical data.
Radius of the Moon's orbit = 3.84 * 108 m
Mass of the Moon = 7.35 * 1022 kg
Time for Moon to complete one orbit around the earth = 2.36 * 106 s
Calculate
(i) the speed of the Moon in its orbit around the Earth,
(ii) the acceleration of the Moon,
(iii) the force the Earth exerts on the Moon,
(iv) the gravitational field strength of the Earth at the Moon.
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Answer:
(a) Gravitational field strength at a point is the gravitational force per unit mass at that point
(b) At Earth's surface, the numerical value of the gravitational field strength is 9.81 ms-2 or 9.81 N/kg (the two units are the same)
(c) It is incorrect because gravity is used for referring to the gravitational force of attraction, mg, exerted on a mass m.
The term g is the acceleration of the mass towards the centre of the earth. It is not gravity.
(d)
Let Rm = radius of Moon's orbit
Mm = mass of Moon
Tm = time for Mon to complete 1 orbit around the Earth
(i) v = Rmω
= Rm * 2π / T
= 2π * (3.84 * 108) / (2.36 * 106)
= 1.022 * 103 ms-1
(ii) acceleration of the Moon = v2 / r
= (1.022 * 103)2 / (3.84 * 108)
= 2.72 * 10-3 ms-2
(iii) The force the Earth exerts on the Moon, F, is given by
F = Mm * gm
= (7.35 * 1022) * (2.72 * 10-3)
= 2.00 * 1020 N
(iv) The gravitational field strength of the Earth at the Moon
= gm
= 2.72 * 10-3 ms-2
