### Q.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 m^{2} kg^{–2}.

**Answer:-**Initial velocity of the rocket, *v* = 2 km/s = 2 × 10^{3} m/s

Mass of Mars, *M* = 6.4 × 10^{23} kg

Radius of Mars, *R* = 3395 km = 3.395 × 10^{6} m

Universal gravitational constant, G = 6.67× 10^{–11} N m^{2} kg^{–2}

Mass of the rocket = *m*

Initial kinetic energy of the rocket = (1/2)*mv*^{2}

Initial potential energy of the rocket = -G*Mm* / *R*

Total initial energy = (1/2)*mv*^{2}– G*Mm* / *R*

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) *mv*^{2} – G*Mm */ *R* = 0.4*mv*^{2} – G*Mm* / *R*

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 = *-G*Mm* / (*R* + *h*)

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

0.4*mv*^{2} – G*Mm* / *R* = -G*Mm* / (*R* + *h*)

0.4*v*^{2} = GM / R – G*M* / (*R* + *h*)

= G*Mh* / *R*(*R* + *h*)

(*R* + *h*) / *h* = G*M* / 0.4*v*^{2}*R*

*R* / *h* = ( G*M* / 0.4*v*^{2}*R* ) – 1

h = R / [ (GM / 0.4v^{2}*R*) – 1 ]

= 0.4*R*^{2}*v*^{2} / (GM – 0.4*v*^{2}R)

= 0.4 × (3.395 × 10^{6})^{2} × (2 × 10^{3})^{2} / [ 6.67 × 10^{-11} × 6.4 × 10^{23} – 0.4 × (2 × 10^{3})^{2} × (3.395 × 10^{6}) ]

= 18.442 × 10^{18} / [ 42.688 × 10^{12} – 5.432 × 10^{12} ]

= 18.442 × 10^{6 }/ 37.256

= 495 × 10^{3} m = 495 km.