First slide

Motion of planets and satellites

Question

A body is projected with a velocity equal to 3/4th of the escape velocity from the surface of the earth. The height it reaches is (Radius of the earth is R)

Easy

A

10R/9
B

9R/7
C

9R/8
D

10R/3

Solution

$v = \frac{3 v_{e}}{4} = \frac{3}{4} \sqrt{\frac{2 G M}{R}} \Rightarrow v^{2} = {(\frac{3}{4} \sqrt{\frac{2 G M}{R}})}^{2} = (\frac{9}{16} \times \frac{2 G M}{R}) = \frac{9 G M}{8 R} - - - (1) T E_{1} = T E_{2} \Rightarrow K E_{1} + P E_{1} = K E_{2} + P E_{2} \Rightarrow \frac{1}{2} m v^{2} - \frac{G m M}{R} = 0 - \frac{G m M}{R + h} \Rightarrow \frac{1}{2} m (\frac{9 G M}{8 R}) - \frac{G m M}{R} = 0 - \frac{G m M}{R + h} \Rightarrow \frac{9}{16 R} - \frac{1}{R} = - \frac{1}{R + h} \Rightarrow \frac{9 - 16}{16 R} = - \frac{1}{R + h} \Rightarrow \frac{- 7}{16 R} = - \frac{1}{R + h} \Rightarrow 7 R + 7 h = 16 R \Rightarrow h = \frac{16 R - 7 R}{7} = \frac{9 R}{7}$

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