# 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 - Physics

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.

#### Solution 1

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 = 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

Concept: Gravitational Potential Energy
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Chapter 8: Gravitation - Exercises [Page 202]

#### APPEARS IN

NCERT Class 11 Physics
Chapter 8 Gravitation
Exercises | Q 25 | Page 202

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