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MCQ · Q4

Q.The binding energy of a satellite revolving around a planet in a circular orbit is 3×1093\times10^9 J. Its kinetic energy is (A) 6×1096\times10^9 J (B) −3×109-3\times10^9 J (C) −6×109-6\times10^9 J (D) 3×1093\times10^9 J

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For a satellite of mass m in a stable circular orbit of radius r, KE=12GMmrKE=\dfrac{1}{2}\dfrac{GMm}{r} and PE=−GMmrPE=-\dfrac{GMm}{r}, so the total energy is TE=KE+PE=−12GMmr=−KETE=KE+PE=-\dfrac{1}{2}\dfrac{GMm}{r}=-KE. Binding energy is defined as the energy needed to just free the satellite, BE=−TEBE=-TE. Combining these two relations, BE=−TE=−(−KE)=KEBE=-TE=-(-KE)=KE -- kinetic energy and binding energy are always numerically EQUAL for a circular orbit. Given BE=3×109BE=3\times10^9 J, it follows directly that KE=3×109KE=3\times10^9 J. (T …

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