JPJC 2026 Gravitational Field Assignment with solutions

2026/JPJC/PHYSICS/9478 1 JURONG PIONEER JUNIOR COLLEGE 9478 H2 PHYSICS GRAVITATIONAL FIELD ASSIGNMENT 1 A geostationary satellite of mass 1500 kg orbits the Earth in a circle of radius r with period T. It is known that the two quantities are related by the equation nT Ar= . The Earth has a mass of 246.0 10 kg and a radius of 66.4 10 m. (a) Determine the values of n and A. Include appropriate units with your values. [4] (b) Calculate the distance of the orbit from the surface of the Earth. [2] (c) Calculate the total energy of the satellite in orbit. [2] 15

2026/JPJC/PHYSICS/9478 2 (d) Determine the energy required to put this satellite into orbit from the equator, taking into consideration the rotation speed of the Earth. [2] (e) As the satellite orbits the Earth, it gradually loses energy because of air resistance. (i) With reference to the total energy of the satellite, state and explain the effect of this change on the radius of the orbit. [2] ……………………………………………………………………………………………….. ……………………………………………………………………………………………….. ……………………………………………………………………………………………….. (ii) Suggest why so many satellites “burn up” in the Earth’s atmosphere. [3] ………………………………………………………………………………………………. ………………………………………………………………………………………………. ………………………………………………………………………………………………. ……………………………………………………………………………………………….

2026/JPJC/PHYSICS/9478 3 JURONG PIONEER JUNIOR COLLEGE 9748 H2 PHYSICS GRAVITATIONAL FIELD ASSIGNMENT SOLUTIONS No. Solution Remarks 1 (a) Gravitational force provides centripetal force 2 2 2 2 2 3 3 2 3 2 4 4 4 gcFF GMm mr r GM r T Tr GM GM r    = = = = = 3 2n = 73.14 10A −= 3 2sm − [1] for correct statement [1] for correct substitution [1] correct A with correct units [1] correct n with no units [Deduct one mark overall for any wrong unit.] 1 (b) 2 3Tr A= 2 3 7 7 24 3600 3.1408 10 4.23 10 m − =  = 76Distance from surface of Earth 4.23 10 6.4 1 0=  −  73.6 10 m= [1] for correct radius of orbit [1] for correct distance 1 (c) Total energy of satellite = GPE + KE 2 2 11 24 7 9 9 1 2 1 2 2 6.67 10 (6.0 10 )(1500) 2(4.2287 10 ) 7.096 10 J 7.1 10 J GMmmv r GM GMmm rr GMm r − = + −  =−  =− =−  =−  −  [1] for correct substitution [1] for correct answer

2026/JPJC/PHYSICS/9478 4 1 (d) Energy at equator of Earth = GPE + KE   − =− + =− + =− + =−  +  =−  +  − 2 Earth 2 Earth 2 Earth 11 24 6 62 10 8 10 1 2 1 ()2 12 ()2 6.67 10 (6.0 10 )(1500) (6.4 10 ) 1 2(3.1416)(1500)[(6.4 10 ) ]2 (24 3600) 9.3797 10 1.6246 10 J = 9.3634 10 J Earth Earth GMm mvr GMm mrr GMm mrrT Energy required to put the satellite into orbit =−  − − 9 107.096 10 ( 9.3634 10 ) =  10 10 8.6538 10 J 8.7 10 J [1] for correct energy at surface of Earth [1] for correct answer 1 (e) (i) As the satellite loses total energy, it becomes more negative. Hence, the magnitude of 2 GMm r increases, then r decreases. Hence, the radius of orbit decreases. [1] for explanation [1] for conclusion 1 (e) (ii) As r decreases, 2 GMm r increases. Hence, KE increases and speed increases. There is more air resistance (due to higher speed), and the satellites fall faster and faster towards the Earth with a smaller and smaller radius, leading to more heat generated until the satellite burns up. [1] KE increases [1] speed increases [1] conclusion of speeding up and smaller radius