Question

22 Example 10.53 Calculate the escape velocity from the surface of a planet of mass 14.8 x 10 kg. It is given that radius of the e planet is 3.48 x 10⁶ m. 22​

Answer 1

Answer:

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Explanation:

$to \: find \: : \\ escape \: velocity \: of \: planet \: (Vp) \\ \\ given : \\ \\ mass \: of \: planet \: (M) = 14.8 \times 10 {}^{22} \: kg \\ \\ radius \: of \: planet \: (R) = 3.48 \times 10 {}^{6} \: m \\ \\ G = 6.67 \times 10 {}^{ - 11} \: \frac{N.m {}^{2} }{kg {}^{2} }$

$so \: we \: know \: that \\ \\ escape \: velocity = \sqrt{ \frac{2GM}{R} } \\ \\ thus \: then \: accordingly \\ \\ Vp = \sqrt{ \frac{2GM}{R} } \\ \\ = \sqrt{ \frac{2 \times 6.67 \times 10 {}^{ - 11} \times 14.8 \times 10 {}^{22} }{3.48 \times 10 {}^{6} } } \\ \\ = \sqrt{ \frac{2 \times 6.67 \times 14.8 \times 10 {}^{ - 11} \times 10 {}^{22} \times 10 {}^{ - 6} }{3.48} } \\ \\ = \sqrt{ \frac{197.432 \times 10 {}^{ - 11 + 22 - 6} }{3.48} } \\ \\ = \sqrt{ \frac{197.432 \times 10 {}^{5} }{3.48} } \\ \\ = \sqrt{56.733 \times 10 \times 10 {}^{4} } \\ \\ = \sqrt{567 .33 \times 10 {}^{4} } \\ \\ = 23.8186 \times 10 {}^{2} \\ \\ = 2.38186 \times 10 \times 10 {}^{2} \\ \\ Vp = 2.38186 \times 10 {}^{3} \: \: \frac{km}{s}$