RI Y5 Remedial G-field Soln

Raffles Institution Physics Department Y5 H2 Physics Term 3 Remedial 4 Chapter 7: Gravitational Field Solutions 1 A  2 34 3 2 4 3 GMg r Gr r G r      Hence, when density is constant, g is proportional to r (or d). 2 D GM r  1 2 1800 400 kJ kg2 Q P PQ Q r X rX         Work done on 1-kg mass by gravitational force from P to Q = initial PE  final PE = 800  (400) = 400 kJ Remarks: 1. The negative answer makes sense because the gravitational force on the mass acts towards the planet, which is in the opposite direction to the displacement from P to Q. 2. It is useful to remember that work done by gravitational force = loss in GPE and loss in GPE = initial PE  final PE. 3 B From 2 GMmF r : Gravitational force decreases with a larger radius. From GMmU r : GPE becomes less negative and hence increases with a larger radius. From 2 2 we have mv GMm GM,vrr r : Linear speed decreases with a larger radius. From 3 GM r  : Angular velocity decreases with a larger radius. 4 D Since gravitational force on the satellite provides the centripetal force, cag 2 22 4 2 6 2 4 3 2 3 2 R g M M gr T rT g T RT g TT        

Raffles Institution Physics Department Y5 H2 Physics Term 3 Remedial 5 5 (a) 51 2 2 24 3600 7.27 10 rad s T        M1 A1 (b) 75 31 4.23 10 7.27 10 3.08 10 m s vr        M1 A1 (c) 2 75 2 4.23 10 7.27 10 0.226 m s ar        M1 A1 (d) Gravitational attraction provides the centripetal force: 2400 0.226 537 N gFm a   M1 A1 (e)  2 2 27 11 24 537 4.23 10 6.67 10 2400 6.0 10 kg g g GMmF r FrM Gm        M1 A1 6 (a) (i) Gravitational field strength is numerical ly equal to the potential gradient at the point. The direction of the gravitational field strength is towards lower potential. B1 B1 (ii) The potential at infinity is defined to be zero. Since gravitational force is attractive in nature, to bring a mass from infinity to a point in the gravitational field, the direction of the external force is opposite to the direction of displacement of the mass. This results in negative work done by the external force. B1 B1 (b) (i)  6720 10 2.0 4.0 10 JppEm      A1 (ii)   6 6 1 10 34 10 Gravitational field stength 30 5 10 0.96 N kg      M1 A1 (iii) 1Difference in potential 40 MJ kg  A1

Raffles Institution Physics Department Y5 H2 Physics Term 3 Remedial 6 (iv) To escape, the total energy 0TE  2 Earth 0 1 02 PKEE GMm mvR    Earth 2GMv R  Earth 2escape velocity GM R M1 A1 7 (a) (i) 51 51 2angular velocity 24 3600 7.272 10 rad s 7.27 10 rad s        B1 (ii)  2 265 2 centripetal acceleration 6.38 10 7.27 10 0.0337 m s R        M1 A1 (b) (i) An object at the equator is undergoing circular motion with radius 6.38  106 m. The effective acceleration of free fall is thus given by the difference between the gravitational field strength and the centripetal acceleration: 2eff c GMga R At the poles, the radius of the circular motion is zero. Hence ac = 0 and 2eff GMg R B1 B1 (ii) The difference is ac which has a value of 0.0337 m s2 which is small compared to 9.81 m s2. A1