Problem 79 - Master Resource Book in JEE Main Physics

Question

The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is \(v\). For a satellite orbiting at an altitude of half of the earth's radius, the orbital velocity is (a) \(\frac{3}{2} v\) (b) \(\sqrt{\frac{3}{2}} v\) (c) \(\sqrt{\frac{2}{3}} v\) (d) \(\frac{2}{3} v\)

Step-by-Step Solution

Answer

The correct answer is (c) \(\sqrt{\frac{2}{3}} v\).

Step 1: Understanding the Problem

We need to find the new orbital velocity of a satellite orbiting at an altitude of half of Earth's radius, given that its orbital velocity when it is just above Earth's surface is \( v \). We will use the formula for orbital velocity to find the new velocity.

Step 2: Recalling the Orbital Velocity Formula

The formula for orbital velocity for a satellite in a circular orbit is given by \[ v = \sqrt{\frac{GM}{r}} \]where \( G \) is the gravitational constant, \( M \) is the mass of the Earth and \( r \) is the orbital radius.

Step 3: Orbital Radius for Each Orbit

When the satellite is just above the Earth's surface, the orbital radius \( r \) is equal to the Earth's radius \( R \). For the satellite at an altitude of \( \frac{R}{2} \), the new orbital radius \( r' = R + \frac{R}{2} = \frac{3R}{2} \).

Step 4: Substitute Radii into the Formula

Now calculate the orbital velocity at the new altitude:\[v' = \sqrt{\frac{GM}{\frac{3R}{2}}}\]By simplifying, \[v' = \sqrt{\frac{2GM}{3R}} \]

Step 5: Comparing with Initial Velocity

We know that when the satellite is just above Earth's surface, \( v = \sqrt{\frac{GM}{R}} \). To compare \( v' \) to \( v \), substitute \( v \) into \( v' \):\[v' = \sqrt{\frac{2}{3}} v \]Thus, \( v' = \sqrt{\frac{2}{3}}v \).

Step 6: Select the Correct Answer

From the provided options, the correct one that matches our derived result is (c) \(\sqrt{\frac{2}{3}} v\).

Key Concepts

Gravitational Constant

The gravitational constant, denoted as \( G \), is a fundamental physical constant that plays a crucial role in the law of universal gravitation. This constant is essentially a measure of the force of gravity between two objects. The value of \( G \) is approximately \( 6.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \). It helps us calculate the attractive force between two masses in space. This force is what keeps planets orbiting stars, moons orbiting planets, and satellites orbiting Earth.

When dealing with orbital velocity calculations, \( G \) is paired with the mass of the celestial body (like Earth) to determine how a satellite will behave in its orbit. It's interesting to note that while gravity pulls us down towards Earth, it also keeps the Moon in orbit around us and the Earth in orbit around the Sun.

Circular Orbit

A circular orbit occurs when a satellite or celestial body revolves around another body in a path that forms a perfect circle. This is a special type of orbit characterized by constant speed and an orbital path where the distance from the orbiting object to the central body remains constant.

• In physics, circular orbits are significant as they allow us to better predict the behavior of satellites and plan their launch and operations.
• Satellites in circular orbits tend to maintain a consistent altitude and speed, making them easier to manage and maintain.

For objects in circular orbit, the formula for orbital velocity, \( v = \sqrt{\frac{GM}{r}} \), is used to determine the speed needed to remain in this orbit. Balancing gravitational pull and the inertia of the satellite's motion determines this speed. Thus, a circular orbit is not just a neat path but a precise balance of cosmic forces.

Orbital Radius

The orbital radius is an essential component in understanding how satellites and celestial bodies maintain their orbits. It is the distance from the center of the central body, such as Earth, to the orbiting satellite. For satellites, the orbital radius is crucial to determine their speed and stability in orbit.

In the exercise example, the initial orbital radius is equivalent to Earth's radius \( R \). When the satellite moves to a higher altitude, the new orbital radius becomes \( \frac{3R}{2} \), reflecting both the altitude and the distance from Earth's center.

• Alterations in the orbital radius require recalculations of speed using the orbital velocity formula.
• The greater the orbital radius, the slower the orbital speed, due to the increased path length that the satellite must cover.

Understanding the relationship between orbital radius and velocity helps in satellite placement and ensuring they remain in stable orbits without veering off path.