JPJC 2026 Gravitational Field Tutorial Solutions

1 JURONG PIONEER JUNIOR COLLEGE 9478 H2 PHYSICS GRAVITATIONAL FIELD TUTORIAL SOLUTIONS Gravitational constant, 11 2 2106.67 N m kgG −−=  Self-Check Questions S1 Gravitational field strength at a point is defined as the gravitational force per unit mass acting at that point. S2 Consider a mass M and another mass m placed at point X, which is at a distance r away from mass M. S3 Gravitational potential at a point is the work done per unit mass by an external force in bringing a small test mass from infinity to that point. S4 A geostationary orbit refers to a circular orbit on which satellite is fixed about a position above the Earth. (1) Period is 24 hours, (2) Orbit is above equator, (3) Satellite moves from west to east direction.

2 Self Practice Questions P1 (a) 9 11 106.125 )30)(20)(1067.6( −− =F N (b) 9106.1 − N (c) (d) The gravitational force that mass A exerts on mass B is the same magnitude but opposite direction as the gravitational force that mass B exerts on mass A. These forces are of the same type and act on two different bodies. Hence, gravitational forces are action-reaction pairs. P2 2 -11 24 62 -1 (6.6730 x 10 )(5.9768 x 10 ) (6.3782 x 10 ) 9.8038 N kg = = = GMg r P3 (a) 10100.45 )0.1)(30( −−−=−= GrGMmU J (b) Gravitational potential at a point is of the same value as gravitational potential energy of a 1 kg mass at the same point in a gravitational field. However, these are two different physical quantities. Gravitational potential is a quantity associated at every point around a source mass , whereas gravitational potential energy is a property of system of 2 masses separated by a distance . The answer in (a) has the unit Joules, which is different as the units of gravitational potential, Joules per kilogram. (c) 10100.4 − J (d) The negative sign shows that gravitational force is attractive. P4 A Only gravitational force FG acts on the satellite, which provides the centripetal force Fc to maintain the satellite in circular motion. P5 Work done on the unit mass by the external force = Final potential – initial potential 33 ()2 2 2GM GM GM GM GMR R R R R= − − − = − = − 30 kg 20 kg

3 Discussion Questions 1 Gravitational force on the satellite by the Earth, Fg provides the centripetal force, Fc to keep the satellite in orbit. 2 2 2 11 24 2 7 7 (6.67 )(6.0 ) (2500) 6.403.. 10 10 10 10 m 6.4 m gcFF GMm mv rr GMr v − = = = = =     2 (a) −− −−= −   11 11 10 1(6.67 10 )(30) (6.67 10 )(30) 4.2 10 N kg4 25g (b) (c) (i) magnitude increases to 1.42 x 10─9 Nkg─1 but direction remains unchanged. −− −−= −   11 11 91(6.67 10 )(90) (6.67 10 )(30) 1.42 10 N kg4 25g (ii) magnitude decreases but direction changes to the left. −− −−= −   11 11 11 1(6.67 10 )(30) (6.67 10 )(30) 9.7 10 N kg9 16g (iii) magnitude will increase. Direction will change. −− −− − −− −− = −   == = + = 11 11 10 1 11 10 1 2 2 10 1 (6.67 10 )(30) (6.67 10 )(30) 4.2 10 N kg4 25 (6.67 10 )(30) 5.0 10 N kg4 6.5 10 N kg x y xy g gx g g g x -10 y -10 x o g 5.0x10tanθ = =g 4.2x10 θ=49.9 above horizontal, pointing top right. 30 kg point X 30 kg

4 3 , 11() () PQ PQ QP PQ QP PQ GM GM rr GM rr rrGM rr     =− =−  = − =− −= Both potential will increase in value. The difference will decrease, as seen from the graph. 4 (a) (i)  −=−+ =− 2 GM GM rr GM r (ii)    − +− =− = sin sin 2 sin GM GM rr GM r (b) (i) 2 2 2 2cos cos cosGM GM rr g GM r    == + (ii) 0 (oppositely directed). (c) Yes, it is possible. If the vector sum is zero, it is possible for the scalar sum to be non- zero because all values for potential are negative. 5 (a) The gravitational force on satellite by the Earth provides the centripetal force to keep the satellite in orbit. cg FF = 2 2 =GMm mvRR 21 22 = GMmmv R 2=K GMmE R (b) GMmU r=− Hence, 2 2 GPE GMm r KE r GMm −=  =−

5 (c) (i) Fig. 5 10.0 8.0 6.0 4.0 2.0 0 −2.0 −4.0 −6.0 −8.0 −10.0 0 PR 2 PR 3 PR 4 PR R / m energy / 109 J

6 (ii) At 4 PRR= , 9 9 9 1 2.5 J KE 1.25 J2 2(KE) 2(1.25 J)v 1600 1250m 10 10 1 s 0 P i i U E m − =− ==− = = =   At 2 PRR= , 9 9 9 1 1 5 J KE 2.5 J2 2(KE) 2(2.5 J)v 1600 1767.76...ms 1700ms 10 10 10 f P f U E m − − =− ==− == =     Hence, 1 1770 1250 520 m s fi vvv − = − = − = 6 (a)(i)  = = =T -522 7.27×1024×60×60 rad s−1 (a)(ii) 2 =a = r 6 -5 2(6.38×10 )(7.27×10 ) = 0.0337 m s−2 (b) (i) Centripetal force is provided by the gravitational force, 3 3 2 2 3 2 2 2 22 2 2 3 3 6 5 7 6.389.81 7.27 4.23 m 10 10 10 GMm mr r GMr GMr GM R R Rg      − = = = =   =      =   =

7 (ii) Both the Earth and the satellite must be rotating in the same direction (west to east). The satellite must be directly above the equator (must lie in the same plane as the equator) so that it shares the same axis of rotation as the Earth. Note: (Optional: advantages of geostationary orbits and polar orbits can be discussed.) Many systems, such as the Global Positioning System, use several satellites in low orbits that pass over the Earth’s poles. Suggest an advantage of these low polar orbits, and an advantage of geostationary orbits. Advantage for polar orbits: • There are no fixed number of satellites that can be launched into orbit. • the distance is small so signal travels quickly. Advantage for geostationary orbits: • Continuous transmission occurs between satellite and ground station. • Less energy is required to launch satellites into orbit. 7 (a) The gradient of the graph represents resultant gravitational field str ength ( dr dg −= ). Maximum on the graph and hence zero gradient on the graph occurs when the gravitational field strength due to the Moon and gravitational field strength due to the Earth cancels each other. The resultant gravitational field strength is zero. The curve is asymmetrical as the mass of the Earth is l arger than the mass of the Moon. The gravitational potential GM r =− on the surface of the M oon and on the Earth are different and hence gravitational pull of the Earth is stronger compared to that of the Moon. (b) The resultant force can be derived obtaining the negative of the gradient of the r− graph, and multiplying it by the mass of the tektite. If the value is negative, it indicates that the force is directed (radially inwards) towards the moon. If the value is positive, it indicates that the force is directed (radially outwards), towards the Earth. (c) ME FF = 11.022 == E MEM M M y x y mGM x mGM (d) The tektite must be able to reach a point P in order to be attracted by the field of the Earth. Otherwise, it falls back to the Moon. From the graph, At the surface of the Moon,  = 6-3.9×10 J 4 PRkg−1 At P, 6103.1 −= J 4 PRkg−1 Using conservation of energy, K.E.M + P.E.M = K.E.P + P.E.P

8 ( ) ( ) 2280103.10109.321 0662 −+=−+ VmmmVo m 4 PRs−1 (e) Total energy at the surface of moon = Total energy at the surface of Earth 1 2 1 2 2 1 2 1  mmVmmV oo +=+ ( )1 22 1 2  −=− ooVV From graph, 11oo VV   (where o is the gravitational potential at the surface of moon, 1 is the gravitational potential at the surface of Earth) A tektite will reach the Earth at a speed greater than oV . 8 [2016/P3/Q9] (a)(i) 2P GMg x = (magnitude) (a)(ii) P GMmE x=− (b)(i)

9 (b)(ii) Gravitational field strength is directly proportional to the mass of the sphere. A lower mass will r