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}.

#### Solution 1

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 = `1/2mv^2`

Initial potential energy of the rocket = =`(-Gmm)/R`

Total initial energy = `1/2mv^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 xx 1/2mv^2 - (GMm)/R = 0.4 mv^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)`

`=GM(1/R - 1/(R+h))`

`=GM((R+h-R)/(R(R+h)))`

`=(GMh)/(R(R+h))`

`(R+h)/h = (GM)/(0.4v^2R)`

`R/h +1 = (GM)/(0.4v^2R)`

`R/h = (GM)/(0.4v^2R) -1`

`h =R/((GM)/(0.4v^2R)) -1`

`= (0.4R^2v^2)/(GM-0.4v^2R)`

=`(0.4xx(3.395xx10^6)^2xx(2xx10^3)^2)/(6.67xx10^(-11)xx6.4xx10^23-0.4xx(2xx10^3)^2xx(3.395xx10^6))`

=`(18.442xx10^(18))/(42.688xx10^(12)-5.432xx10^(12))` = `18.442/37.256 xx10^6`

`=495xx10^3 m = 495 km`

#### Solution 2

Initial K.E = `1/2mv^2`;Initial P.E = -`(GMm)/R`

Where m= Mass of rocket M= Mass of Mars, R = Radius of Mars.

∴Total initial energy = `1/2mv^2 - (GMm)/R`

Since 20% of k.E is lost,only 80% remain to reach th height

∴Total inital energy available

`=4/5xx1/2 mv^2 - (GMm)/R`

`=0.4mv^2 - (GMm)/R`

When the rocket reaches the highest point at a height h above the surface its K.E is zero and P.E = -`(GMm)/(R+h)`

Using principle of conversion of energy

`0.4 mv^2 - (GMm)/R = (GMm)/(R+h)`

or `(GMm)/(R+h)= (GMm)/R - 0.4mv^2 => (GM)/(R+h) = (GM)/R - 0.4 v^2`

or `(GM)/(R+h) = 1/R[GM-0.4 Rv^2] => (R+h)/R = (GM)/(GM-0.4 Rv^2)`

or `h/R = (GM)/(GM-0.4Rv^2) - 1`

or `h/R = (0.4Rv^2)/(Gm-0.4Rv^2) => h = (0.4R^2v^2)/(GM-0.4 Rv^2)`

or `h = (0.4xx(2xx10^3)^2xx(3.395xx10^6)^2)/(6.67xx10^(-11)xx6.4xx10^(23)-0.4xx(2xx10^3)^2xx3.395xx10^6)` m

=495km