__Question 8.25__:^{-11}N m

^{2}kg

^{–2}.

__Solution__:*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 = -GMm / R

Total initial energy = (1/2)mv

^{2}- GMm / 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) X (1/2) mv

^{2}- GMm / R = 0.4mv

^{2}- GMm / 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 =*-GMm / (R + h)

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

0.4mv

^{2}- GMm / R = -GMm / (R + h)

0.4v

^{2}= GM / R - GM / (R + h)

= GMh / R(R + h)

(R + h) / h = GM / 0.4v

^{2}R

R / h = ( GM / 0.4v

^{2}R ) - 1

h = R / [ (GM / 0.4v

^{2}R) - 1 ]

= 0.4R

^{2}v

^{2}/ (GM - 0.4v

^{2}R)

= 0.4 X (3.395 X 10

^{6})

^{2}X (2 X 10

^{3})

^{2}/ [ 6.67 X 10

^{-11}X 6.4 X 10

^{23}- 0.4 X (2 X 10

^{3})

^{2}X (3.395 X 10

^{6}) ]

= 18.442 X 10

^{18}/ [ 42.688 X 10

^{12}- 5.432 X 10

^{12}]

= 18.442 X 10

^{6 }/ 37.256

= 495 X 10

^{3}m = 495 km

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