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 = 6.4×10^{23} kg; radius of mars = 3395 km; G = 6.67×10^{-11} N m^{2} kg^{–2}.

Asked by Pragya Singh | 1 year ago | 153

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 × 10^{6} 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 – 0.4Rv^{2}]

(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

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