Question

The gravitational potential at height 'h'above earth's
surface is -5.12 x10'J/kg and acceleration due to
gravity at this point is 6.4 ms 2. If the radius of earth
is 6400 km, the value of his​

Answer 1

SOLUTION

$gravity \: at \: a \: height \: h = g {(1 + \frac{h}{r} })^{ - 2} \\ where \: r \: is \: the \: radius \: of \: the \: earth \\ g \: is \: the \: gravity \: at \: surface \: of \: earth \\ \\ acc \: to \: you \: gravity \: at \: h = 6.4m {sec}^{ - 2} \\ g {(1 + \frac{h}{r} })^{ - 2} = 6.4 \\ 10 {(1 + \frac{h}{r}) }^{ - 2} = 6.4 \\ {(1 + \frac{h}{r}) }^{ - 2} = 0.64 \\ {(1 + \frac{h}{r} )}^{2} = \frac{100}{64} \\ {( \frac{r + h}{r}) }^{2} = \frac{100}{64} \\ \\ \frac{6400000 + h}{6400000} = \frac{10}{8} \\ 8(6400000 + h) = 10(6400000) \\ 8h = 2(6400000)m \\ \\ h = 1600km......(1) \\ \\ potential \: energy \: at \: height \: h \\ - \frac{GmM}{h} \\ \\ acc \: to \: you \: potential \: energy \: = - 5.12 \times 10 \\ \\ \frac{GmM}{h} = 5.12 \times 10 \\ \\ \frac{6 \times {10}^{24} \times 6.6 \times {10}^{ - 11} \times m }{1600} =5.12 \times 10 \\ \\ 0.24 \times {10}^{13} \times m = 5.12 \times 10 \\ m = \frac{5.12 \times 10}{0.24 \times {10}^{13} } \\ \\ m = 21.3\times {10}^{ - 12} kg$

Answer 2

Explanation:

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