Problem 781 - JEE Physics

Question

A satellite with K.E. \(E_{\mathrm{k}}\) is revolving round the earth in a circular orbit. How much more K.E. should be given to it so that it may just escape into outer space ? (A) \(\mathrm{E}_{\mathrm{k}}\) (B) \(2 \mathrm{E}_{\mathrm{k}}\) (C) \((1 / 2) \mathrm{E}_{\mathrm{k}}\) (D) \(3 \mathrm{E}_{\mathrm{k}}\)

Step-by-Step Solution

Answer

The additional kinetic energy required for the satellite to escape Earth's gravity is equal to its current kinetic energy (\(E_{k}\)). Therefore, the correct answer is (A) \(E_{k}\).

Step 1: Find escape velocity formula

The formula to calculate escape velocity is given by: \(v_{e} = \sqrt{\frac{2GM}{R}}\), where \(v_{e}\) is the escape velocity, \(G\) is the gravitational constant, \(M\) is the mass of the Earth, and \(R\) is the distance from the center of the Earth to the satellite, which is the sum of Earth's radius and the altitude of the satellite.

Step 2: Find orbital velocity formula

The formula to calculate orbital velocity is given by: \(v_{o} = \sqrt{\frac{GM}{R}}\), where \(v_{o}\) is the orbital velocity.

Step 3: Calculate kinetic energies using velocities

The kinetic energy of the satellite in its current state is given by \(E_{k} = \frac{1}{2}mv_{o}^2\), and the kinetic energy it needs to have to escape Earth's gravity is given by \(E_{k_{e}} = \frac{1}{2}mv_{e}^2\).

Step 4: Substitute the expressions for velocities

Now, substitute the expressions for \(v_{o}\) and \(v_{e}\) from steps 1 and 2 into the kinetic energy equations: \(E_{k} = \frac{1}{2}m\left(\sqrt{\frac{GM}{R}}\right)^2\) \(E_{k_{e}} = \frac{1}{2}m\left(\sqrt{\frac{2GM}{R}}\right)^2\)

Step 5: Simplify the expressions for kinetic energies

Upon simplification, we get: \(E_{k} = \frac{1}{2}m\left(\frac{GM}{R}\right)\) \(E_{k_{e}} = \frac{1}{2}m\left(\frac{2GM}{R}\right)\)

Step 6: Find the additional kinetic energy required

The additional kinetic energy required is the difference between \(E_{k_{e}}\) and \(E_{k}\): Additional Kinetic Energy = \(E_{k_{e}} - E_{k}\) = \(\frac{1}{2}m\left(\frac{2GM}{R}\right) - \frac{1}{2}m\left(\frac{GM}{R}\right)\) By simplifying the equation, we get: Additional Kinetic Energy = \(E_{k}\) So, the correct answer is (A) \(E_{k}\).

Key Concepts

Escape Velocity

When talking about escape velocity, we are trying to understand the speed needed for an object to break free from a planet's gravitational pull. To achieve this, an object must travel at a certain minimum velocity. This is what we call the escape velocity.
For Earth, the escape velocity can be calculated using the formula: \[ v_{e} = \sqrt{\frac{2GM}{R}} \] Where,

• \( G \) is the universal gravitational constant,
• \( M \) is the mass of Earth,
• \( R \) is the distance from the center of Earth to the object.

Escape velocity doesn't depend on the mass or size of the object trying to escape, only on Earth's mass and the distance from its center. It is essentially the tipping point speed an object needs to never return. Whenever a satellite wants to leave Earth's orbit and fly into space, it will need to hit this velocity as the minimum speed.

Orbital Velocity

Orbital velocity is the speed needed for an object to stay in a stable orbit around a planet. Imagine it as a balance between the object's speed and the gravitational pull of the planet.
Without enough speed, the object would fall back to Earth. At the right speed, which is the orbital velocity, the object will stay in a consistent orbit. This is expressed by the equation: \[ v_{o} = \sqrt{\frac{GM}{R}} \] This formula has similar components to the escape velocity equation but favors a stable orbit instead of breaking free.

• Just like with escape velocity, \( G \), \( M \), and \( R \) refer to the gravitational constant, Earth's mass, and the distance from its center respectively.

Objects in orbit move faster the closer they are to Earth's surface. Orbital velocity ensures that a satellite keeps circling the Earth without floating away or crashing down.

Kinetic Energy of Satellites

Kinetic energy is a measure of the motion energy an object possesses while in motion. For satellites, this energy varies depending on what is needed: staying in orbit or escaping into space.
The kinetic energy for an orbiting satellite is calculated with its orbital velocity: \[ E_{k} = \frac{1}{2} m v_{o}^2 \] Here,

• \( m \) is the mass of the satellite,
• \( v_{o} \) is its orbital velocity.

In contrast, to allow the satellite to escape Earth, the energy must be calculated with the escape velocity: \[ E_{k_{e}} = \frac{1}{2} m v_{e}^2 \] In order to shift from orbiting to escaping, a satellite must increase its kinetic energy. It turns out, as shown in the simplified solution, to break away completely, the satellite requires an additional kinetic energy equal to the current kinetic energy it already holds in orbit.

Gravitational Constant

The gravitational constant, denoted as \( G \), is a key factor in calculating gravitational forces. It is a universal constant that appears prominently in the formulas for escape velocity and orbital velocity.
The value of \( G \) is approximately \[ G = 6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \] \( G \) plays a crucial role by linking the mass of two objects and the force of attraction between them, which in turn affects both how orbits are calculated and how much speed is required for escaping Earth's pull.
Knowing \( G \) allows us to understand and quantify the strength of gravity, not just on Earth, but anywhere in the universe. This fundamental constant makes our calculations of movement in the cosmos consistent and predictable.