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.

Asked by Pragya Singh | 1 year ago |  153

Solution :-

Speed of the rocket fired from the surface of mars ( v) = 2 km/s

Mass of the rocket = m

Mass of the Mars ( M) = 6.4 × 10²³ Kg

Radius of the Mars (R) = 3395 km = 3.395 × 106 m

Initial Potential energy =$$\dfrac{– GMm}{R}$$

Initial kinetic energy of the rocket = $$\dfrac{1}{2}$$ mv²

Total initial energy = $$\dfrac{1}{2}$$mv² –$$\dfrac{GMm}{R}$$

Due to atmospheric resistance, 20% of the kinetic energy is lost by the rocket.

The remaining Kinetic energy= 80% of $$\dfrac{1}{2}$$mv²

$$\dfrac{2}{5}$$ mv²= 0.4 mv² ——–(1)

When the rocket reaches the highest point, at the height h above the surface the kinetic energy will be zero and the potential energy is equal to

$$\dfrac{– GMm}{R+h}$$

Applying Law of conservation of energy.

$$\dfrac{– GMm}{R+h}$$ = $$\dfrac{– GMm}{R}$$ + (0.4) mv²

$$\dfrac{– GMm}{(R+h)}$$ = $$\dfrac{1}{R}$$ [ GM – 0.4Rv2]

(R+$$\dfrac{h}{R}$$) = $$\dfrac{GM}{(GM – 0.4Rv^2)}$$

$$\dfrac{h}{R}$$ = $$\dfrac{GM}{(GM – 0.4Rv^2)}$$ – 1

$$\dfrac{h}{R}$$$$\dfrac{0.4Rv^2}{(GM – 0.4Rv^2)}$$

h =$$\dfrac{0.4R^2v^2}{(GM – 0.4Rv^2)}$$

h = $$\dfrac{0.4\times (2\times 10^{3})\times (3.395\times 10^{6})^{2}}{6.67\times 10^{-11}\times 6.4\times 10^{23}-0.4\times (2\times 10^{3})^{2}\times 3.395\times 10^{6}}$$

h = 495 km

Answered by Abhisek | 1 year ago

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