Notes : Derivation Total Energy and Binding Energy of an Orbiting Satellite | Param Himalaya

Tuesday, June 16, 2026

Param Himalaya Logo

Notes : Derivation Total Energy and Binding Energy of an Orbiting Satellite

Objective: To derive expressions for the Total Energy and Binding Energy of an orbiting satellite.


Total Energy of a Satellite

The total energy of a satellite in its orbit is the sum of its Potential Energy (P.E.) and Kinetic Energy (K.E.).

$$E = U + K.E.$$

Where,

  • $M$ = Mass of Earth
  • $m$ = Mass of satellite
  • $R$ = Radius of Earth
  • $h$ = Height of satellite above Earth's surface
  • $r = R+h$ = Radius of orbit

1. Potential Energy of Satellite

The gravitational potential energy of a satellite revolving around the Earth in a circular orbit is:

$$ U = -\frac{GMm}{R+h} $$

or

$$U = -\frac{GMm}{r} $$

2. Kinetic Energy of Satellite

The kinetic energy of the satellite is:

$$ K.E.=\frac{1}{2}mv_0^2 $$

where $v_0$ is the orbital velocity.

Since the satellite moves in a circular orbit,

Centripetal Force = Gravitational Force

$$ \frac{mv_0^2}{R+h} = \frac{GMm}{(R+h)^2} $$

or

$$ mv_0^2=\frac{GMm}{R+h} $$

Substituting in the kinetic energy equation,

$$K.E. = \frac{1}{2} \left( \frac{GMm}{R+h} \right) $$
 $$ K.E. = \frac{GMm}{2(R+h)} $$

or

$$ K.E. = \frac{GMm}{2r} $$

3. Total Energy of Satellite

$$ E = U + K.E. $$
$$ E = -\frac{GMm}{R+h} + \frac{GMm}{2(R+h)} $$
$$ E = -\frac{GMm}{2(R+h)} $$

Since

$$ r=R+h $$

therefore,

$$ E = -\frac{GMm}{2r} $$

Significance

Significance

If E<0, the satellite remains bound to Earth.

If E=0, the satellite escapes Earth's gravitational field.

If E>0, the satellite moves away from Earth.


Binding Energy of a Satellite

Definition

The minimum energy required by a satellite to escape from its orbit to infinity is called the Binding Energy.

Energy of satellite in orbit:

$$ E=-\frac{GMm}{2r} $$

At infinity,

$$ E_{\infty}=0 $$

Therefore,

$$ B.E. = 0-E $$
$$ B.E. = 0- \left( -\frac{GMm}{2r} \right) $$
$$B.E. = \frac{GMm}{2r} $$

or

$$ B.E. = \frac{GMm}{2(R+h)} $$

Important Facts

If K.E. of a Satellite is Doubled

Original total energy:

$$E = \frac{GMm}{2(R+h)} - \frac{GMm}{R+h} $$
 $$ E = -\frac{GMm}{2(R+h)} $$

If kinetic energy is doubled,

$$ K.E.' = 2(K.E.) = \frac{GMm}{R+h} $$

Then,

$$ E' = K.E.' + P.E. $$
$$ E' = \frac{GMm}{R+h} - \frac{GMm}{R+h} $$
$$ E'=0 $$

Hence, the satellite escapes from Earth's gravitational field.


Important Relations

$$K.E.=-E $$
 $$ K.E.=-\frac{U}{2} $$
 $$ E=\frac{U}{2} $$

Frequently Asked Questions (FAQ)

Q1. Why is the total energy of a satellite negative?

Because the satellite is gravitationally bound to the Earth.

$E=-\frac{GMm}{2r} $

A negative total energy indicates a bound system.

Q2. What is the relation between K.E. and P.E.?

$ K.E.=-\frac{U}{2} $

Q3. What is the relation between K.E. and Total Energy?

$ K.E.=-E $

Q4. What is Binding Energy?

The minimum energy required to move a satellite from its orbit to infinity.

$ B.E.=\frac{GMm}{2r} $

Q5. What happens if the kinetic energy is doubled?

The total energy becomes zero and the satellite escapes from Earth's gravitational field.

Q6. On what factors does Binding Energy depend?

  • Mass of Earth ($M$)
  • Mass of satellite ($m$)
  • Orbital radius ($r$)
$ B.E.=\frac{GMm}{2r} $

Quiz (MCQs)

  1. The total energy of an orbiting satellite is:
    A. Positive
    B. Zero
    C. Negative
    D. Infinite
    Answer: C. Negative

  2. The kinetic energy of a satellite is:
    A. $U$
    B. $-U$
    C. $-\frac{U}{2}$
    D. $\frac{U}{2}$
    Answer: C

  3. Binding energy of a satellite is:
    A. $\frac{GMm}{r}$
    B. $\frac{GMm}{2r}$
    C. $\frac{GMm}{4r}$
    D. $\frac{2GMm}{r}$
    Answer: B

  4. If total energy becomes zero, the satellite will:
    A. Continue in orbit
    B. Fall towards Earth
    C. Escape from Earth
    D. Stop moving
    Answer: C

  5. SI unit of binding energy is:
    A. Watt
    B. Newton
    C. Joule
    D. Pascal
    Answer: C

Quick Revision Formulae

$$ K.E.=\frac{GMm}{2r} $$ 
$$ U=-\frac{GMm}{r} $$
$$ E=-\frac{GMm}{2r} $$ 
$$ B.E.=\frac{GMm}{2r} $$ 
$$ K.E.=-E $$ $$ K.E.=-\frac{U}{2} $$ 
$$ E=\frac{U}{2} $$ 
$$ r=R+h $$

Conclusion: A satellite remains in orbit because its total energy is negative, and the minimum energy required to free it from Earth's gravitational field is called its binding energy.

No comments:

Post a Comment