Q. 193.9( 12 Votes )

# 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 × 10^{24} kg; radius of the earth = 6.4 × 10^{6} m; G = 6.67 × 10^{–11} N m2 kg^{–2}.

Answer :

Suppose a satellite is orbiting earth’s surface, Seattleite will be out of earth’s influence when its total energy is Zero or positive and we know Total energy of a body is sum of kinetic energy and potential energy

T = K + U

Where T is the total energy, U is potential energy and K is kinetic energy kinetic energy of a body and is always positive, and depends upon the speed of the body, potential energy is negative and decreases as we move away from orbiting the planet and at an infinite distance from the planet

Kinetic Energy is given as

Where K is the kinetic energy of a body of mass m, moving with velocity v

The potential energy of a body above the surface of the earth is given as

U = -GMm/R

Where U is the gravitational Potential Energy of body of mass m at a distance R from the centre of the earth and M is the mass of earth G is universal gravitational Constant

If Body have to move out of Earth’s influence its total energy should be positive i.e.

T ≥ 0

Or

so we need to give it energy which will increase its speed and increase it's kinetic energy and total energy will be positive assume earth to be a sphere of radius r, and satellite is orbiting the earth at a distance h above the surface of the earth at a distance R from the centre of the earth as Shown in figure

So we have

R = r + h

So Total energy T of the satellite orbiting at height h above the surface of the earth will be

T = U_{i} + K_{i}

Where U_{i} and K_{i} are initial kinetic and potential energy of the satellite respectively i.e.

where M is mass of earth, m is mass of satellite, r is Radius of earth, and v_{o} is the orbital velocity of the satellite

The total energy of satellite will be negative and is half of the potential energy, we follow the relation

T = U/2 = -K

Where T is the total energy, U is potential energy and K is kinetic energy

So we can see kinetic energy is (-1/2) times potential energy

So we get

K_{i} = -U/2 = GMm/(2(r+h)

i.e.

Suppose E amount of energy is expended on a rocket to make its total energy 0

i.e.

E + T = 0

Or E = -T

So Energy expended to rocket the satellite out of the earth’s gravitational field will be negative of initial total energy

Where M is mass of earth which is

M = 6.0 × 10^{24} kg

m is mass of satellite which is

m = 200 kg

G is universal gravitational constant

G = 6.67 × 10^{–11} N m2 kg^{–2}

r is the radius of the earth

r = 6.4 × 10^{6} m

h is the initial height of satellite above the surface of the earth

h = 400 Km = 4 × 10^{5} m

so putting values in above equation, we get

So The energy expended to move the satellite out of earth’s gravitational field is 5.9 x 10^{9} J

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