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Solution - 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

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

Reference Material

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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