Problem 801 - JEE Physics

Question

A satellite of mass \(\mathrm{m}\) is circulating around the earth with constant angular velocity. If radius of the orbit is \(\mathrm{R}_{0}\) and mass of earth \(\mathrm{M}\), the angular momentum about the center of earth is (A) \(m \sqrt{\left(G M R_{0}\right)}\) (B) \(\mathrm{M} \sqrt{\left(\mathrm{GMR}_{\mathrm{o}}\right)}\) (C) \(\left.m \sqrt{(G M} / R_{0}\right)\) (D) \(\mathrm{M} \sqrt{\left(\mathrm{GM} / \mathrm{R}_{\mathrm{o}}\right)}\)

Step-by-Step Solution

Answer

The short answer is: The angular momentum of the satellite about the center of Earth is given by \(L = m\sqrt{GMR_0}\), which corresponds to option (A).

Step 1: Find the linear velocity of the satellite

The linear velocity, \(v\), of an object in circular motion can be given by: \(v = ωr\) where \(ω\) is the angular velocity, and \(r\) is the radius of the circle (in this case, the orbit of the satellite).

Step 2: Relate linear velocity to gravitational force

The centripetal force acting on the satellite is equal to the gravitational force acting on it. We can use Newton's law of gravitation to write this force as: \(F = \frac{GmM}{r^2}\) where \(G\) is the gravitational constant, \(m\) is the mass of the satellite, \(M\) is the mass of the Earth, and \(r\) is the radius of the circular orbit (\(R_0\)). In circular motion, centripetal force can also be written as: \(F = \frac{mv^2}{r}\) Equating these two expressions for the centripetal force gives: \(\frac{GmM}{r^2} = \frac{mv^2}{r}\)

Step 3: Solve for linear velocity

We can now solve the equation for the linear velocity, \(v\), as follows: \(v^2 = \frac{GmMr}{r^3}\) \(v^2 = \frac{GM}{r}\) \(v = \sqrt{\frac{GM}{r}}\)

Step 4: Compute the angular momentum

Now that we have the expression for the linear velocity, we can substitute it back into the formula for the angular momentum: \(L = mvr\) \(L = m\cdot\sqrt{\frac{GM}{r}}\cdot r\) Since \(r = R_0\), \(L = m\sqrt{GM R_0}\) Comparing our result with the given answer options, we find that the correct expression for the angular momentum of the satellite is \(L = m\sqrt{GMR_0}\), which corresponds to option (A).

Key Concepts

Centripetal Force

When an object moves in a circle, it constantly changes direction. This change in direction indicates that there is an acceleration present. Any acceleration needs a force, and the force that keeps an object moving in a circle is called the centripetal force. The term "centripetal" comes from Latin, meaning "center-seeking." The centripetal force acts inward, towards the center of the circle. Without this force, the object would move in a straight line.

• Centripetal force formula: \(F = \frac{mv^2}{r}\)
• Where \(m\) is mass, \(v\) is velocity, and \(r\) is radius of the circle.

Interestingly, this force is not a distinct type of force. It is rather the net force required to keep the object in circular motion. In the case of the orbiting satellite, the centripetal force is provided by the gravitational pull of the Earth.

Newton's Law of Gravitation

Isaac Newton proposed the Universal Law of Gravitation to explain the attraction between masses. According to this law, every point mass attracts every other point mass by a force pointing along the line intersecting both points. The magnitude of this force is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

• Gravitational force formula: \(F = \frac{GmM}{r^2}\)
• \(G\) is the gravitational constant, \(m\) and \(M\) are the masses, \(r\) is the distance between the centers of the two masses.

This fundamental force is what keeps planets, moons, and satellites in their respective orbits. In the previous exercise, the gravitational force acts as the centripetal force that keeps the satellite moving in its circular path.

Circular Motion

Circular motion occurs when an object moves along a circular path. This motion can be uniform, meaning the speed is constant, or non-uniform, where the speed changes.In the context of satellites orbiting a planet, they exhibit uniform circular motion. Here, the velocity direction changes continuously, even if the speed remains constant.

• The linear velocity is given by \(v = \omega r\)
• \(\omega\) is angular velocity and \(r\) is the radius of the orbit.

Circular motion is critical in understanding planetary orbits and other phenomena in space. As seen from the exercise, keeping track of forces involved, speed, and direction is essential to find quantities like angular momentum.

Gravitational Constant

The gravitational constant, denoted by \(G\), is a key constant in Newton's Law of Gravitation. It helps quantify the gravitational force between two masses.

• Value of \(G\): \(6.674×10^{-11} \text{Nm}^2/\text{kg}^2\).
• This constant plays a crucial role in calculating forces at astronomical scales.

It's crucial for deriving equations that describe satellite motion and other gravitational phenomena. In terms of the discussed exercise, \(G\) is integral in calculating the gravitational force that acts as the centripetal force, allowing us to derive linear velocities and eventually, angular momentum.