GravitationMedium
Question
If a body of mass 1 kg falls on the earth from infinity, it attains velocity (v) and kinetic energy (k) on reaching the surface of earth. The values of v and k respectively are $\_\_\_\_$ .
(Take radius of earth to be 6400 km and $g = 9.8\text{ }m/s^{2}$ )
Options
A.$11.2\text{ }km/s;6.27 \times 10^{7}\text{ }J$
B.$11.2\text{ }km/s;12.54 \times 10^{7}\text{ }J$
C.$8.8\text{ }km/s;6.27 \times 10^{7}\text{ }J$
D.$8.8\text{ }km/s;12.54 \times 10^{7}\text{ }J$
Solution
Sol. [Key Concept: Energy conservation from infinity: $v = \sqrt{2gR}$ and $k = \frac{1}{2}{mv}^{2}$.]
M.E.C
$${K_{i} + U_{i} = K_{f} + U_{f} }{\Rightarrow O - \frac{{GM}_{e}m}{\infty} = \frac{1}{2}{mv}^{2} - \frac{GMm}{Re} }{V = \sqrt{\frac{2GM}{Re}} = \sqrt{2\text{ }gRe} = \sqrt{2 \times 9.8 \times 6400 \times 1000} }{= 11.2\text{ }km/s }$$Kinetic energy $= \frac{1}{2}{mv}^{2}$
$$= 6.27 \times 10^{7}\text{ }J $$
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