Problem 21 - A Complete Resource Book in Physics for JEE Main

Question

The orbital velocity of an artificial satellite in a circular orbit just above earth's surface is \(v_{0} .\) For a satellite orbiting in a circular orbit at an altitude of half of earth's radius is (A) \(\sqrt{\frac{3}{2}} v_{0}\) (B) \(\sqrt{\frac{2}{3}} v_{0}\) (C) \(\frac{3}{2} v_{0}\) (D) \(\frac{2}{3} v_{0}\)

Step-by-Step Solution

Answer

The orbital velocity of an artificial satellite in a circular orbit at an altitude half of the earth's radius is \(\boxed{\sqrt{\frac{2}{3}} v_{0}}\), which corresponds to option (B).

Step 1: Gravitational force acting on the satellite

Since the satellite is orbiting around the earth, we know that the gravitational force acting on the satellite is responsible for providing the necessary centripetal force for its circular motion. The gravitational force is given by the formula: $$F = G\frac{Mm}{r^2}$$ where \(G\) is the gravitational constant, \(M\) is the mass of the earth, \(m\) is the mass of the satellite, and \(r\) is the distance between the center of the mass of the satellite and the earth.

Step 2: Centripetal force needed for circular motion

The centripetal force required for the satellite to maintain its circular motion is given by the formula: $$F_c = m\frac{v^2}{r}$$ where \(v\) denotes the orbital velocity of the satellite.

Step 3: Equate gravitational force and centripetal force

Since the gravitational force is providing the centripetal force for the satellite's circular motion, we can equate the two formulas from Steps 1 and 2: $$G\frac{Mm}{r^2} = m\frac{v^2}{r}$$

Step 4: Solve for orbital velocity

Now we can solve this equation for the orbital velocity \(v\). First, we simplify the equation by dividing both sides by \(m\), and then multiplying both sides by \(r\): $$G\frac{M}{r} = v^2$$ Then, we take the square root of both sides to obtain the orbital velocity formula: $$v = \sqrt{G\frac{M}{r}}$$

Step 5: Find the orbital velocity at half of the earth's radius

Let's denote the earth's radius as R. The given altitude is half of the earth's radius, therefore the distance from the center of the earth to the satellite will be \(r = R + \frac{R}{2} = \frac{3R}{2}\). Now we can plug this value into our orbital velocity formula derived in step 4: $$v = \sqrt{G\frac{M}{\frac{3R}{2}}}$$ Recall that the orbital velocity just above the earth's surface is given as \(v_0\), so we have: $$v_0 = \sqrt{G\frac{M}{R}}$$ Now, we want to find the ratio of \(v\) to \(v_0\): $$\frac{v}{v_0} = \frac{\sqrt{G\frac{M}{\frac{3R}{2}}}}{\sqrt{G\frac{M}{R}}} = \sqrt{\frac{R}{\frac{3R}{2}}} = \sqrt{\frac{2}{3}}$$ Thus, the orbital velocity of an artificial satellite in a circular orbit at an altitude half of the earth's radius is \(\boxed{\sqrt{\frac{2}{3}} v_{0}}\), which corresponds to option (B).

Key Concepts

Gravitational Force

Gravitational force is the fundamental force that attracts two bodies with mass towards each other. This force is crucial for keeping satellites in orbit around Earth. The formula to calculate gravitational force is given by:

• \( F = G \frac{Mm}{r^2} \)

In this equation,

• \( G \) represents the gravitational constant.
• \( M \) denotes the mass of the Earth.
• \( m \) is the mass of the satellite.
• \( r \) is the distance between the center of the Earth and the satellite.

Gravitational force provides the necessary inward pull that maintains the satellite's trajectory in a stable path. The further away the satellite is, the weaker the gravitational pull becomes due to the square of the distance \( r^2 \) in the denominator. This force is also the reason why objects fall towards the ground on Earth. Every satellite's motion depends on balancing this force.
Remember, gravity is the anchor that ensures satellites don't drift aimlessly into space!

Centripetal Force

Centripetal force is what keeps an object moving in a circular path. When a satellite orbits Earth, the centripetal force is directed towards the center of the circle (which in this case is Earth's center). The formula to determine this force is:

• \( F_c = m \frac{v^2}{r} \)

Here,

• \( F_c \) is the centripetal force required.
• \( m \) is the mass of the satellite.
• \( v \) stands for the satellite's orbital velocity.
• \( r \) is the radius of the circular path (distance from Earth's center).

Centripetal force is essential to combat the inertia that would prefer the satellite to move straight out into space. It is what strings the satellite along its curved orbit, rather like a strong rope pulling the satellite inward. For orbits, this force comes from gravity; allowing a natural alignment between the satellite's speed and the gravitational pull, maintaining a consistent orbit.
This invisible leash keeps satellites zipping along their designated paths!

Circular Motion

Circular motion refers to the movement of an object along the circumference of a circle. In the context of satellites, it describes how the satellite rotates around the Earth. For circular motion to be sustained, the centripetal force must be consistently applied, directing right angles to the object's velocity.
The key properties of circular motion include:

• Uniform circular motion implies constant speed, but the direction changes.
• The orbit is dictated by a stable balance between gravitational pull and inertia.

Satellites experience a unique form of circular motion where the maximum speed is achieved at the closest point to Earth (perigee) due to stronger gravitational pull, and slows down at the farthest point (apogee) slightly, though its orbital period remains stable. An important aspect of this motion is that it reduces atmospheric drag, making the orbit more sustainable. By understanding and controlling the circular motion, satellites can effectively perform tasks like communication, scientific observation, or weather forecasting.
Thanks to circular motion, satellites can tirelessly orbit and gather data from our gorgeous blue planet.