# A Satellite Orbits the Earth at a Height of 400 Km Above the Surface. How Much Energy Must Be Expended to Rocket the Satellite Out of the Earth’S Gravitational Influence? Mass of the Satellite = 200 Kg; Mass of the Earth = 6.0 ×1024 Kg; Radius of the Earth = 6.4 ×106 M - Physics

A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg; mass of the earth = 6.0 ×1024 kg; radius of the earth = 6.4 ×106 m; G = 6.67 × 10–11 N m2 kg–2

#### Solution 1

Mass of the Earth, M = 6.0 × 1024 kg

Mass of the satellite, m = 200 kg

Radius of the Earth, Re = 6.4 × 106 m

Universal gravitational constant, G = 6.67 × 10–11 Nm2kg–2

Height of the satellite, h = 400 km = 4 × 105 m = 0.4 ×106 m

Total energy of the satellite at height h = 1/2 mv^2 + (-GM_em)/(R_e+h)

Orbital velocity of the satellite, v = sqrt("GM"/(R_e+h))

Total energy of height, h = 1/2 m ((Gm_e)/(R_e+h)) - (GM_em)/(R_e+h) = -1/2 ((GM_em)/(R_e+h))

The negative sign indicates that the satellite is bound to the Earth. This is called bound energy of the satellite.

Energy required to send the satellite out of its orbit = – (Bound energy)

=1/2 (GM_em)/(R_e+h)

=1/2 xx (6.67 xx 10^(-11) xx 6.0 xx 10^(24) xx 200)/(6.4xx10^6+0.4xx10^(6))

=1/2 xx (6.67 xx 6 xx 2xx10)/(6.8xx10^6) = 5.9 xx 10^9 J

#### Solution 2

Total energy of orbiting satellite at a height h

=-(GMm)/(R+h) + 1/2 mv^2 = -(GMm)/(R+h) + 1/2m (GM)/(R+h)

=-(GMm)/(2(R+h))

Energy expended to rocket the satellite out of the earth's gravitational field.

=-(total energy of the orbiting satellite)

(GMm)/(2(R+h)) = ((6.67xx10^(-11))xx(6xx10^24)xx200)/(2xx(6.4xx10^6+4xx10^5))

=5.9 xx 10^9 J

Concept: Energy of an Orbiting Satellite
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#### APPEARS IN

NCERT Class 11 Physics
Chapter 8 Gravitation
Q 19 | Page 202