Problem 89 - A Complete Resource Book in Physics for JEE Main

Question

A satellite is revolving in a circular orbit at a height \(h\) from the earth's surface (radius of earth \(R . h \ll R\) ). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is close to: (Neglect the effect of atmosphere) (A) \(\sqrt{g R}\) (B) \(\sqrt{g R / 2}\) (C) \(\sqrt{g R}(\sqrt{2}-1)\) (D) \(\sqrt{2 g R}\)

Step-by-Step Solution

Answer

The minimum increase in orbital velocity required for the satellite to escape from Earth's gravitational field is close to \(\sqrt{gR} (\sqrt{2}-1)\), which corresponds to option (C).

Step 1: Find the initial orbital velocity

We'll first find the orbital velocity of the satellite. The formula to calculate the orbital velocity, \(v_o\), is: \(v_o = \sqrt{\frac{GM}{r}}\) where: G is the gravitational constant (\(6.674 \times 10^{-11} \, m^3 \, kg^{-1} s^{-2}\)) M is the mass of the Earth (\(5.972 \times 10^{24} \, kg\)) r is the distance between the center of the Earth and the satellite Since h ≪ R, we can consider the satellite as moving close to the Earth's surface. Therefore, r in this case is approximately equal to the radius of the Earth (R). \(v_o = \sqrt{\frac{GM}{R}}\) Note that we're not using actual values for G, M, and R but rather their relationship.

Step 2: Find the escape velocity

Next, we need to find the escape velocity, \(v_e\), which is the minimum velocity required for the satellite to escape Earth's gravitational field. The formula for escape velocity is: \(v_e = \sqrt{\frac{2GM}{r}}\) Again, since h ≪ R, we can consider r ≈ R. \(v_e = \sqrt{\frac{2GM}{R}}\)

Step 3: Find the minimum increase in orbital velocity

The minimum increase in orbital velocity, \(\Delta v\), required for the satellite to escape Earth's gravitational field is the difference between the escape velocity and the initial orbital velocity: \(\Delta v = v_e - v_o\) Substitute the values of \(v_e\) and \(v_o\) from Steps 1 and 2: \(\Delta v = \sqrt{\frac{2GM}{R}} - \sqrt{\frac{GM}{R}}\) Now, let's factor out \(\sqrt{GM/R}\): \(\Delta v = \sqrt{GM/R} (\sqrt{2}-1)\) Recall that \(g = \frac{GM}{R^2}\), so we can rewrite the equation as: \(\Delta v = \frac{R\sqrt{g}}{\sqrt{R}} (\sqrt{2}-1)\) Finally, we can simplify this to: \(\Delta v = \sqrt{gR} (\sqrt{2}-1)\) The minimum increase in orbital velocity required for the satellite to escape from Earth's gravitational field is close to \(\sqrt{gR} (\sqrt{2}-1)\), which corresponds to option (C).

Key Concepts

Escape Velocity

Understanding the concept of escape velocity begins with the idea of overcoming a celestial body's gravitational pull. In simple terms, escape velocity is the speed needed for an object to break free from the gravitational field of a planet or any other body in space without further propulsion.

It's important to know that escape velocity doesn't depend on the mass of the object trying to escape but rather on the mass and radius of the celestial body it is escaping from. The formula for escape velocity is given by \[\begin{equation} v_e = \sqrt{\frac{2GM}{r}} \end{equation}\]where:

• \( G \) is the gravitational constant,
• \( M \) is the mass of the celestial body, and
• \( r \) is the distance from the center of the mass to the point of escape.

In the case of a satellite orbiting Earth, the satellite needs a specific minimum increase in its orbital velocity to reach this escape velocity, which will allow it to leave Earth's gravitational influence forever.

Exercise improvement advice for this section recommends an inclusion of examples or problems that demonstrate how the escape velocity changes with varying distances from the Earth's center, emphasizing that as one moves further from Earth, the escape velocity decreases.

Gravitational Field

The concept of a gravitational field is crucial in understanding how objects like satellites move around larger bodies like Earth. This invisible field exudes a force that pulls objects towards the center of the massive body. The strength of the gravitational field is determined by two primary factors: the mass of the object creating the field, and the distance from that object.

The gravitational field strength at a point in space, often denoted by \( g \), is defined by the force exerted per unit mass at that point. It is expressed mathematically as \[\begin{equation}g = \frac{GM}{r^2}\end{equation}\]

• \( G \) again being the gravitational constant,
• \( M \) the mass of the celestial body creating the field,
• and \( r \) the distance between the object's center and the point where the field strength is measured.

A satellite in orbit is constantly 'falling' into this gravitational field but has enough horizontal velocity to keep missing Earth, thus maintaining its orbit. The solution provided in the exercise essentially makes use of Earth's gravitational field strength to calculate the necessary orbital velocities.

The advice for improving exercise content here would be to explore the variations of the gravitational field at different altitudes, thereby showing how it affects the orbital velocity of satellites at varying heights.

Circular Orbit

A circular orbit is a particular path in which satellites or celestial bodies travel around a bigger body due to the force of gravity. For an object to maintain a stable circular orbit, it must balance the gravitational force pulling it towards the larger body with its tendency to move in a straight line, known as inertia. When these forces balance out just right, the object moves in a circular path at a constant speed, known as the 'orbital velocity'.

The formula for the orbital velocity (\( v_o \)) of a satellite close to Earth's surface is given by\[\begin{equation} v_o = \sqrt{\frac{GM}{r}}\end{equation}\]It's important to note that in a perfectly circular orbit, the orbital velocity remains constant. This is due to the fact that the gravitational force acting as the centripetal force needed for circular motion is consistent at a given altitude.

In the provided exercise, the step-by-step solution calculates the orbital velocity and then finds the additional speed required for the satellite to transition from a circular orbit to escaping Earth's gravity.

Including interactive visual representations of circular orbits, and the balance of gravitational force and inertia could greatly enhance students' grasp of this concept, as recommended by the exercise improvement advice.