Q. 193.9( 12 Votes )

# A satellite orbit

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 = Ui + Ki

Where Ui and Ki 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 vo 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

Ki = -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 × 1024 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 × 106 m

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

h = 400 Km = 4 × 105 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 109 J

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses RELATED QUESTIONS :

A satellite is inPhysics - Exemplar

A satellite is toPhysics - Exemplar

<span lang="EN-USHC Verma - Concepts of Physics Part 1

<span lang="EN-USHC Verma - Concepts of Physics Part 1

<span lang="EN-USHC Verma - Concepts of Physics Part 1

As you have learnNCERT - Physics Part-I

What is the anglePhysics - Exemplar

<span lang="EN-USHC Verma - Concepts of Physics Part 1

A satellite orbitNCERT - Physics Part-I

Satellites orbitiPhysics - Exemplar