## Escape velocity and orbital velocity of satellite (Derivation, example and Q and N)

**Escape Velocity**

**Orbital velocity**

**Time period and height of Satellite**

**Geostationary satellite and parking orbit**

**Energy of sattellite**

**Escape velocity **

It is the velocity with which any object thrown toward the atmosphere goes beyond the earth's gravitational field. It means it is the minimum velocity required to throw any object beyond the earth's gravitational field.

The name behind the escape is velocity which is the minimum velocity needed to escape from the earth's gravitational field.

To calculate the earth's escape velocity, let us consider the earth having mass (M_{e}) and radius (R). we have to throw any object of mass (m) from earth surface (A) so that it goes beyond the earth's gravitational field.

Let us consider a body having mass (m) is at a point B in the gravitational field. Then the gravitational force of attraction between the earth and object of mass (m) is given by

Now amount of work done in moving object from B to C through distance dx is given by

W= F_{g}.dx (work done W= F×d)

Now, the total amount of work done in moving the body from A to out of earth gravitational field is given by

This is the total amount of work done in moving objects from the earth's surface (point A) to the out of the earth's gravitational field.

Let V_{e} be the escape velocity of the earth the amount of work done in moving object from earth surface (potential energy) should be equal to kinetic energy i.e.

W=K.E. … (1)

Where

And

From equation (1) we get

∴

Or * *

In case of earth

Acceleration due to gravity g= 9.8 m/s^{2}

Radius of earth = 6400 km =6400×10^{3}

Then scape velocity of the earth is given by

**Orbital velocity **

Orbital velocity can be defined as the velocity required by a satellite to revolve around the earth into its orbit. This is the minimum velocity satellite should have to revolve around the earth in a particular orbit.

To calculate the orbital velocity of the earth's satellite, let us consider the earth having mass M_{e} and radius R as shown in figures. We have to calculate the orbital velocity of a satellite having mass m which is revolving around the earth at height h from the earth's surface and at a distance r from the center of the earth. Let V_{0} be the orbital velocity of the satellite then the centripetal force required to revolve around the earth is given by

* * … (1)

Where r=R+h

Now, the gravitational force of attraction between the earth and satellite is given by

… (2)

The satellite will be revolved in balanced mode only when the centripetal force required and gravitational force are equal to each other hence from equation (1) and equation (2) we get

F_{c}=F_{g}

Or,

Or,

Or,

We know

Then

** **

Which is the expression for the orbital velocity of the earth revolving around the earth.

If the satellite is very close to the earth, the height of the satellite can be neglected hence our expression becomes

In this case velocity of satellite become

**The time period of the satellite **

It is the time taken by a satellite to make complete one that revolves around the earth into its orbit.

We know

* *

* *

* *

Where r=R+h

Hence

** **

** **

Which is a required expression for a time period of satellite. If a satellite is very close to the earth surface, the height of the satellite can be neglected then expression becomes

We know Radius of earth R= 6400 km = 6400×1000m

Acceleration due to gravity g = 9.8 m/sec

Then

Time period

∴ T= 500 sec

**Height of satellite **

It is the distance between the earth's surface and the satellite.

We know the time period of the satellite

Squaring both sides, we get

** **

we know

Hence

** **

which is the height of the satellite.

**Geostationary satellite **

Any satellite which appears to be stationary with respect to the earth is called a geostationary satellite. It is because the time period of the earth and satellite become the same and appears as rest or stationery. The geostationary satellite time period is the same as earth i.e., 24 hours.

The orbit of the geostationary satellite is called the parking orbit.

The height of the geostationary satellite can be calculated from the time period formula

** **

R= 6400 km

g= 9.8 m/s^{2}

T= 24×3600 sec

Then height

** **

∴ *h=36000km*

Which is the height of the geostationary satellite.

**Energy of satellite**

The energy of the satellite is the sum of kinetic energy and potential energy. Kinetic energy is created due to the motion of the satellite and potential energy is created due to the gravitational force of attraction between satellite and earth.

Hence total energy of the satellite is the sum of the kinetic energy and potential energy which can be written as the

The energy of satellite E= K.E + P.E. … (1)

Now, We know gravitational potential energy is given by

… (2)

Where M_{e} = mass of earth

m= mass of the satellite

R= radius of the earth

h= height of the satellite

again, we know kinetic energy is

Where V_{0} is the orbital velocity

We know,

Hence K.E.

∴ … (3)

Now from equation (1), (2), and (3) we get

E= K.E + P.E.

On solving we get

Which is the total energy of the satellite. Negative signs indicate that the satellite is bounded to earth i.e., to take out the satellite from orbit this much energy is required.

- VIA
- Edubomb

- SOURCS
- Edubomb

## LEAVE A COMMENT

## Facebook Conversations