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.

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Anonymous · Accepted Answer

Initial velocity of the rocket, v = 2 km/s = 2 × 10³ m/s

Mass of Mars, M = 6.4 × 10²³ kg

Radius of Mars, R = 3395 km = 3.395 × 10⁶ m

Universal gravitational constant, G = 6.67× 10^–11 N m² kg^–2
Mass of the rocket = m

Initial kinetic energy of the rocket =

mv²
Initial potential energy of the rocket =

Total initial energy =

mv^{2 -}

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

Total initial energy available =

= (80/100) × (1/2) mv² -

= 0.4mv² -

Maximum height reached by the rocket = h

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

Total energy of the rocket at height h =

Applying the law of conservation of energy for the rocket,

0.4mv² -

=

0.4v² =

=

h =

=

= 495 × 10³ m

= 495 km.

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