Introduction

An artificial satellite is a man-made object that is intentionally placed into orbit around a celestial body. In the case of an artificial satellite revolving around the Earth in a circular orbit, its velocity is half the escape velocity. In this response, we will discuss why the height of the satellite from the Earth's surface is.

Escape Velocity

Escape velocity is the minimum velocity an object needs to escape the gravitational pull of a celestial body. For Earth, the escape velocity is approximately 11.2 km/s. This means that an object traveling at or above this velocity can leave Earth's orbit and venture into space.

Velocity of the Artificial Satellite

Given that the velocity of the artificial satellite is half the escape velocity, we can calculate its value. If we denote the escape velocity as V_escape, then the velocity of the satellite (V_satellite) can be expressed as:

V_satellite = 0.5 * V_escape

This means that the satellite is traveling at half the speed required to escape Earth's gravitational pull.

Centripetal Force

For an object in circular motion, there must be a force acting towards the center of the circle to keep it in that orbit. In the case of the artificial satellite, this force is provided by the gravitational force between the satellite and the Earth.

The centripetal force (F_c) can be calculated using the following equation:

F_c = (m * v^2) / r

Where:
- m is the mass of the satellite
- v is the velocity of the satellite
- r is the distance between the satellite and the center of the Earth

Height from Earth's Surface

To determine the height of the satellite from the Earth's surface, we need to consider the radius of the Earth (R). The distance between the satellite and the center of the Earth (r) can be expressed as:

r = R + h

Where:
- R is the radius of the Earth
- h is the height of the satellite from the Earth's surface

Since the satellite is revolving in a circular orbit, the centripetal force is equal to the gravitational force between the satellite and the Earth. We can equate these forces and solve for the height of the satellite (h):

(m * v^2) / r = (G * m * M) / r^2

Where:
- G is the gravitational constant
- M is the mass of the Earth

Simplifying the equation, we get:

v^2 = (G * M) / r

Substituting the value of r as R + h, we have:

v^2 = (G * M) / (R + h)

Since we know that v = 0.5 * V_escape, we can substitute this value and solve for h:

(0.5 * V_escape)^2 = (G * M) / (R + h)

Simplifying further, we arrive at:

h = R * (1 - (0.25 * V_escape^2) / (G * M))

Conclusion

Therefore, the height of the artificial satellite from the Earth's surface can be calculated using the equation mentioned above. By substituting the known values of the escape velocity, gravitational constant, and the mass and radius of the Earth, we can determine the precise height at which the satellite is

The orbital speed = √ GM / D

G = the universal gravitational constant,
M = Mass of Earth
D = Distance of Object from center of Earth

Escape Velocity = √ 2GM / R

The orbital speed = (1/2) Escape Velocity
√ GM / D  =  (1/2) √ 2GM / R

Squaring both sides
GM/D  = (1/4) * 2GM/R
D = 2R

height of the satellite above the surface of the earth is = D - R
= 2R - R
= R