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# Solution for A Rocket is Fired Vertically with a Speed of 5 Km S–1 From the Earth’S Surface. How Far from the Earth Does the Rocket Go before Returning to the Earth? Mass of the Earth = 6.0 × 1024 Kg; Mean Radius of the Earth = 6.4 × 106 M - CBSE (Science) Class 11 - Physics

ConceptGravitational Potential Energy

#### Question

A rocket is fired vertically with a speed of 5 km s–1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 × 1024 kg; mean radius of the earth = 6.4 × 106 m; G= 6.67 × 10–11 N m2 kg2

#### Solution 1

8 × 106 m from the centre of the Earth

Velocity of the rocket, v = 5 km/s = 5 × 103 m/s

Mass of the Earth, M_e = 6.0 xx 10^(24) kg

Radius of the Earth,R_e = 6.4 xx 10^6 m

Height reached by rocket mass, h

At the surface of the Earth,

Total energy of the rocket = Kinetic energy + Potential energ

=1/2 mv^2 + ((-GM_em)/R_e)

At highest point h

v = 0

And Potential energy = ((GM_em)/(R_e+h))

Total energy of the rocket = 0 + ((-GM_em)/(R_e+h)) = -((Gm_em)/(R_e+h))

From the law of conservation of energy, we have

Total energy of the rocket at the Earth’s surface = Total energy at height h

1/2 mv^2 + ((-GM_em)/R_e) = - (GM_em)/((R_e+h)

1/2v^2 = GM_e(1/R_e - 1/(R_e+h))

=GM_e((R_e+h-R_e)/(R_e(R_e+h)))

1/2v^2 = (GM_eh)/(R_e(R_e+h))xxR_e/R_e

1/2xxv^2 = (gR_eh)/(R_e+h)

Where g = GM/R_e^2 = 9.8 m/s^2 (Acceleration due to gravity on the Earth's surfarce)

:.v^2(R_e+h) = 2gR_eh

v^2R_e = h(2gR_e - v^2)

h = (R_ev^2)/(2gR_e-v^2)

=(6.4 xx 10^6 xx (5xx10^3)^2)/(2xx9.8xx6.4xx10^6-(5xx10^3)^2)

h = (6.4xx25xx10^(12))/(100.44xx10^6) = 1.6 xx 10^6 m

Height achieved by the rocket with respect to the centre of the Earth

=R_e+h

=(6.4xx10^6+1.6xx10^6)

=8.0xx10^6 m

#### Solution 2

Initial kinetic energy of rocket = 1/2 mv2 = 1/2 x m x (5000)2 = 1.25 x 107 mJ
At distance r from centre of earth, kinetic energy becomes zero

∴Change in kinetic energy = 1.25 x 107 – 0 = 1.25 x 107 m J
This energy changes into potential energy.

Initial potential energy at the surface of earth = GMem/’r

=(-(6.67xx10^(-11))xx(6xx10^(24))m)/(6.4xx10^6) = -6.25xx10^7 J

Final potential energy at distance r = -(GM_em)/r

=(-(6.67xx10^(-11))xx(6xx10^(24))m)/r = -4xx10^(14) m/r J

:.Change in potential energy= 6.25 xx 10^7 m - 4 xx 10^(14) m/r

Using law of conservatin of energy

6.25xx 10^7 m - (4xx10^14 m)/r =         1.25xx 10^7 m

i.e r = (4xx10^(14))/(5xx10^7) m  = 8xx10^(16) m

Is there an error in this question or solution?

#### APPEARS IN

NCERT Physics Textbook for Class 11 Part 1 (with solutions)
Chapter 8: Gravitation
Q: 17 | Page no. 202
Solution for question: A Rocket is Fired Vertically with a Speed of 5 Km S–1 From the Earth’S Surface. How Far from the Earth Does the Rocket Go before Returning to the Earth? Mass of the Earth = 6.0 × 1024 Kg; Mean Radius of the Earth = 6.4 × 106 M concept: Gravitational Potential Energy. For the course CBSE (Science)
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