Problem 770 - JEE Physics

Question

If the height of a satellite from the earth is negligible in comparison of the radius of the earth \(\mathrm{R}\), the orbital velocity of the satellite is (A) \(\mathrm{gR}\) (B) \((\mathrm{gR} / 2)\) (C) \(\sqrt{(g} / \mathrm{R})\) (D) \(\sqrt{(g R)}\)

Step-by-Step Solution

Answer

The orbital velocity of the satellite can be found by equating the gravitational force and the required centripetal force to keep it in orbit. This results in the equation \(v^2 = gR\). Solving for orbital velocity, we get \(v = \sqrt{gR}\), which corresponds to answer (D).

Step 1: Identify centripetal force and gravitational force

Let's denote the mass of the satellite as m, the Earth's gravitational constant as g, and the Earth's radius as R. The centripetal force acting on the satellite is given by: \[F_c = \frac{mv^2}{R}\] Where v is the orbital velocity and R is the distance between the Earth's center and the satellite (which is still the Earth's radius). The gravitational force acting on the satellite is given by: \[F_g = mg\]

Step 2: Equate the two forces

In order for the satellite to stay in orbit, the gravitational force must be equal to the centripetal force. Therefore: \[\frac{mv^2}{R} = mg\]

Step 3: Solve for orbital velocity, v

Simplify the equation and solve for v. First, divide both sides by m: \[\frac{v^2}{R} = g\] Now, multiply by R to isolate v^2: \[v^2 = gR\] Finally, take the square root of both sides to get v: \[v = \sqrt{gR}\] Therefore, the orbital velocity of the satellite is given by answer (D): \[v = \sqrt{(g R)}\]

Key Concepts

Centripetal Force

In the context of satellite motion, centripetal force is pivotal. When a satellite orbits the Earth, it moves in a circular path. For it to maintain this path without flying off into space or falling to the ground, a constant force is required. This is where centripetal force comes into play. The formula for centripetal force is:

• \( F_c = \frac{mv^2}{R} \)

Here, \( m \) is the mass of the satellite, \( v \) is the orbital velocity, and \( R \) is the radius of the circular path (Earth's radius in this scenario).
This formula essentially tells us how much force is needed to keep the satellite in orbit given its mass, velocity, and orbital radius.
Centipetal force is essential to balance an orbiting satellite so it stays on its path.

Gravitational Force

Gravitational force is the natural phenomenon by which all things with mass are brought toward one another. For satellites orbiting the Earth, this is the key force that pulls them towards the planet. In the context of our satellite problem, the gravitational force acting on the satellite is given by:

• \( F_g = mg \)

Where \( m \) is the mass of the satellite and \( g \) is the acceleration due to Earth's gravity (approximately \( 9.8\, \text{m/s}^2 \) near the surface).
The gravitational force acts towards the center of the Earth, providing the necessary pull that keeps the satellite in orbit.
Essentially, without gravitational force, the satellite would move off in a straight line into space.

Satellite Motion

Satellite motion refers to the path and dynamics of a satellite when it orbits a planet like Earth. To stay in orbit, a satellite must strike a balance between speed and gravitational pull. Too fast, and it might escape into space; too slow, and it would fall back to Earth.
In our scenario, maintaining an accurate orbital velocity is crucial. This velocity allows the satellite to balance centripetal and gravitational forces, effectively enabling it to "fall around" the Earth continuously.
The formula for the required orbital velocity, given negligible height above the surface, is:

• \( v = \sqrt{gR} \)

Where \( g \) is the gravitational constant and \( R \) is the Earth's radius.
This velocity ensures the satellite maintains a stable orbit as it balances the forces acting on it.

Earth's Gravity

Earth's gravity is the invisible force that pulls objects towards the planet's center. It is this force that dictates the motion of satellites and affects everything from the way we walk to how planets orbit the sun.
Gravity provides the necessary force to keep satellites in motion around Earth. It is represented by the standard acceleration due to gravity, \( g = 9.8\, \text{m/s}^2 \), a value used to calculate the gravitational force on orbiting satellites.
This force ensures that satellites do not drift into space but instead maintain a circular orbit above the Earth's atmosphere, reliant on the delicate balance of gravitational and centripetal forces.
Understanding Earth's gravity is key to predicting and controlling satellite trajectories.