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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 - CBSE (Science) Class 11 - Physics

#### Question

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

#### 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)

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#### APPEARS IN

Solution 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 Concept: Escape Speed.
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