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The Escape Speed of a Projectile on the Earth’S Surface is 11.2 Km S–1. a Body is Projected Out with Thrice this Speed. What is the Speed of the Body Far Away from the Earth? Ignore the Presence of the Sun and Other Planets - Physics

The escape speed of a projectile on the earth’s surface is 11.2 km s–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets

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Solution 1

Escape velocity of a projectile from the Earth, vesc = 11.2 km/s

Projection velocity of the projectile, vp = 3vesc

Mass of the projectile = m

Velocity of the projectile far away from the Earth = vf

Total energy of the projectile on the Earth = `1/2mv_p^2 - 1/2mv_"esc"^2`

Gravitational potential energy of the projectile far away from the Earth is zero.

Total energy of the projectile far away from the Earth = `1/2mv_f^2`

From the law of conservation of energy, we have

`1/2mv_p^2 - 1/2mv_"esc"^2 = 1/2 mv_f^2`

`v_f = sqrt(v_p^2-v_"esc"^2)`

`=sqrt((3v_"esc")^2 - (v_"esc")^2`

`=sqrt8 v_"esc"`

`=sqrt8 xx11.2 = 31.68` km/s

Solution 2

Let v_"es" be the escape speed from surfce of Earth havinf a vlaue `v_"es"` = 11.2 kg  `s^(-1)`

=`11.2 xx 10^3 ms^(-1)`

By definition

`1/2 mv_e^2 = (GMm)/(R^2)`

When a body is projected with aspeed `v_i = 3v_"es" = 3 xx 11.2 xx 10^3` m/s then it will have a final spee `v_f` such that

`1/2 mv_f^2 = 1/2mv_i^2 - (GMm)/R^2 = 1/2mv_i^2 - 1/2mv_e^2`

=>`v_f = sqrt(v_i^2 -v_e^2)`

`=sqrt((3xx11.2xx10^3)-(11.2xx10^3)^2)`

= `11.2 xx 10^3 xx sqrt8`

=`31.7xx10^3 ms^(-1)` or 31.7 km `s^(-1)`

 

Concept: Escape Speed
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APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 8 Gravitation
Q 18 | Page 202
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