The following questions on satellites are simple, but slightly different from the common direct questions. See whether you can solve them yourself. Go through the solution given, only after your trial.

**(1)A satellite is moving in an orbit of radius ‘r’ around the earth. A second satellite is moving in an orbit of radius (1.02)r. Then the orbital period of the second satellite is greater than that of the first by approximately**

(a) 0.2% (b) 2% (c) 3% (d) 10.2% (e) 6%

We have T(a) 0.2% (b) 2% (c) 3% (d) 10.2% (e) 6%

^{2}α r^{3}(Kepler’s law) from which T α r^{3/2}. Since the increment in the orbital radius ‘r’ is 0.02r, the fractional increment in the orbital period ‘T’ is (∆T)/T = (3/2)(∆r)/r = (3/2) × 0.02r/r = 0.03. The percentage increase is 0.03×100 = 3% [option (c)].**(2) If the orbital period of a satellite is ‘T’, its kinetic energy is directly proportional to**

(a) T (b) T(a) T (b) T

^{-1}(c) T^{3/2}(d) T^{2/3}(e) T^{-2/3}

We have, kinetic energy, E = GMm/2r where G is the gravitational constant, M is the mass of the earth ‘m’ is the mass of the satellite and ‘r’ is the radius of the orbit. Therefore, E α 1/r. Since T

^{2}α r^{3}in accordance with Kepler’s law, we can write r α T^{2/3}. Therefore, E α T^{-2/3}. The correct option is (e).
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