8.25 A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s–1. If 20% of its initial energy is lost due to Martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of mars = 6.4×10²³ kg; radius of mars = 3395 km; G = 6.67×10^-11 N m² kg^–2.

2 Views | Posted 5 months ago

Answered by

Vishal Baghel | Contributor-Level 10

5 months ago

Initial kinetic energy of the rocket = $\frac{1}{2} {m v}^{2}$

Initial potential energy of the rocket = $\frac{- G M m}{R}$

Total initial energy = $\frac{1}{2} {m v}^{2} - \frac{G M m}{R}$

If 20% of initial kinetic energy is lost due to Martian atmosphere resistance, then only 80% of its kinetic energy helps in reaching a height

Total initial energy available = 0.8 $* \frac{1}{2} {m v}^{2}$ $- \frac{G M m}{R}$

Maximum height reached by the rocket = h

At this height, the velocity and hence the kinetic energy of the rocket becomes zero.

Total energy of the rocket at height h = $- \frac{G M m}{R + h}$

Applying the law of conservation of energy for the rocket, we can write:

0.4 $* {m v}^{2}$ $- \frac{G M m}{R}$ = $- \frac{G M m}{R + h}$

0.4 $v^{2}$ = GM( $\frac{1}{R}$ $- \frac{1}{R + h})$

0.4 $v^{2}$ = GM( $\frac{R + h - R}{R (R + h)}$ )

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Initial kinetic energy of the rocket = $\frac{1}{2} {m v}^{2}$

Initial potential energy of the rocket = $\frac{- G M m}{R}$

Total initial energy = $\frac{1}{2} {m v}^{2} - \frac{G M m}{R}$

If 20% of initial kinetic energy is lost due to Martian atmosphere resistance, then only 80% of its kinetic energy helps in reaching a height

Total initial energy available = 0.8 $* \frac{1}{2} {m v}^{2}$ $- \frac{G M m}{R}$

Maximum height reached by the rocket = h

At this height, the velocity and hence the kinetic energy of the rocket becomes zero.

Total energy of the rocket at height h = $- \frac{G M m}{R + h}$

Applying the law of conservation of energy for the rocket, we can write:

0.4 $* {m v}^{2}$ $- \frac{G M m}{R}$ = $- \frac{G M m}{R + h}$

0.4 $v^{2}$ = GM( $\frac{1}{R}$ $- \frac{1}{R + h})$

0.4 $v^{2}$ = GM( $\frac{R + h - R}{R (R + h)}$ ) = $\frac{G M h}{R (R + h)}$

$\frac{R + h}{h}$ = $\frac{G M}{{0.4 R v}^{2}}$

$\frac{R}{h}$ + 1 = $\frac{G M}{{0.4 R v}^{2}}$ = $\frac{6.4 * 10^{23} * 6.67 * 10^{- 11}}{0.4 * 3395 * 10^{3} * 2000^{2}}$

h = 495 km

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