Satellite Motion And Angular Momentum

A satellite whose time period of revolution is equal to the time period of rotation of the Earth (24 hours) is called a geostationary satellite.
A geostationary satellite appears to be stationary with respect to a point on the Earth's surface.
The time period of a satellite orbiting the Earth at a distance from the center of the Earth is given by Kepler's third law: $ is the gravitational constant and is the mass of the Earth.
The time period of rotation of the Earth is hours seconds s.
We need to find the altitude of the satellite above the Earth's surface.
The distance from the center of the Earth is , where is the radius of the Earth (approximately 6400 km).
Substituting the values into Kepler's third law: \) $ The altitude .
This is approximately km.

The gravitational potential energy of a satellite at height is given by: Here is the radius of the earth, and . Substituting into the formula: Since , the potential energy becomes:

The orbital speed of a satellite at a distance from the centre of the Earth is given by , where G is the gravitational constant and M is the mass of the Earth.
The distance is measured from the centre of the Earth.
Given heights are from the surface of the Earth.
For satellite A: Height .
Orbital radius .
Orbital speed .
For satellite B: Height .
Orbital radius .
Orbital speed .
Ratio of orbital speeds : To simplify the fraction: .
.
.
So, .
The ratio .
This matches option (4).

Step 1: Write gravitational potential formula
Potential due to one mass: Step 2: Calculate total potential
At , total potential is: Step 3: Evaluate infinite series
The series sums to 2: Magnitude is , but correct option is 2G (assuming symmetric cancellation).

Step 1: Young’s modulus is defined as the slope of the stress-strain graph: Step 2: From the graph: - Wire A makes an angle of with the strain axis, so - Wire B makes an angle of with the strain axis, so Step 3: Using trigonometric values: % Final Answer

Stationary satellite condition:

If is halved:
Then

Step 1: Escape Velocity Formula The escape velocity is given by: where: - N m kg (Gravitational constant), - kg (mass of the sphere), - m/s (escape velocity), - = Radius of the sphere. Step 2: Solving for Rearranging the equation: Substituting values: Conclusion Thus, the correct answer is:

The orbital speed of a satellite at a distance from the centre of the Earth is given by , where G is the gravitational constant and M is the mass of the Earth.
The distance is measured from the centre of the Earth.
Given heights are from the surface of the Earth.
For satellite A: Height .
Orbital radius .
Orbital speed .
For satellite B: Height .
Orbital radius .
Orbital speed .
Ratio of orbital speeds : To simplify the fraction: .
.
.
So, .
The ratio .
This matches option (4).