Q#7 (Past Exam Paper – June 2015 Paper 42 Q1)
(a) The Earth may be considered to be a uniform sphere of radius 6.37 × 103 km with its mass of 5.98 × 1024 kg concentrated at its centre. The Earth spins on its axis with a period of 24.0 hours.
(i) A stone of mass 2.50 kg rests on the Earth’s surface at the Equator.
1. Calculate, using Newton’s law of gravitation, the gravitational force on the stone. [2]
2. Determine the force required to maintain the stone in its circular path. [2]
(ii) The stone is now hung from a newton-meter.
Use your answers in (i) to determine the reading on the meter. Give your answer to three significant figures. [2]
(b) A satellite is orbiting the Earth. For an astronaut in the satellite, his sensation of weight is caused by the contact force from his surroundings.
The astronaut reports that he is ‘weightless’, despite being in the Earth’s gravitational field.
Suggest what is meant by the astronaut reporting that he is ‘weightless’. [3]
Solution:
(a)
(i)
1.
Gravitational force F = Gm1m2 / x2
F = (6.67×10–11 × 2.50 × 5.98×1024) / (6.37×106)2
F = 24.6 N
2.
Centripetal force F = mxω2
{Period T = 24 days = 24×3600 s
Angular frequency ω = 2π / T}
Centripetal force F = 2.50 × 6.37×106 × (2π / 24×3600)2
Centripetal force F = 0.0842 N
(ii)
{The centripetal force is the resultant force on the stone.
Forces on the stone when it is hung: weight and tension in spring of newton-meter. The tension in the spring determines the reading on the meter.
Centripetal force = Weight – Tension
Tension = Weight – Centripetal force}
Reading = 24.575 – 0.0842
Reading = 24.5 N
(b)
The gravitational force provides the centripetal force. When the gravitational force FG is ‘equal’ to the centripetal force FC, the ‘weight’ / sensation of weight / contact force / reaction force which is the difference between FG and FC is zero.
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