Problem 7 - Master Resource Book in JEE Main Physics

Question

A spaceship is launched into a circular orbit close to earth's surface. The additional velocity that should be imparted to the spaceship in the orbit to overcome the gravitational pull is (Radius of earth \(=6400 \mathrm{~km}\) and \(g=9.8 \mathrm{~ms}^{-2}\) ) (a) \(11.2 \mathrm{kms}^{-1}\) (b) \(8 \mathrm{kms}^{-1}\) (c) \(3.2 \mathrm{kms}^{-1}\) (d) \(1.5 \mathrm{kms}^{-1}\)

Step-by-Step Solution

Answer

The additional velocity needed is (b) 8 kms^{-1}.

Step 1: Understanding Orbital Velocity

The orbital velocity is the speed needed for an object to stay in a stable orbit at a particular height above Earth. It's determined by the balance between gravitational pull and the inertia of the object's motion.

Step 2: Formula for Orbital Velocity

The formula for calculating orbital velocity (v_o) is \( v_o = \sqrt{\frac{GM}{R}} \), where \( G \) is the gravitational constant \(6.67\times10^{-11} \text{Nm}^2/\text{kg}^2\), \( M \) is the mass of Earth \(6\times10^{24} \text{kg} \), and \( R \) is the radius of Earth \(6.4 \times 10^6 \text{m} \).

Step 3: Calculation of Orbital Velocity

Plug the known values into the formula: \( v_o = \sqrt{\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{6.4 \times 10^{6}}} \). Simplifying gives: \( v_o = \sqrt{\frac{4 \times 10^{14}}{6.4 \times 10^6}} \approx 7.9 \text{ km/s} \), which can be approximated to \(8 \text{ km/s}\) for practical purposes.

Step 4: Answer Selection

Based on the calculations, the additional velocity required for a spaceship to maintain a stable orbit very close to the Earth's surface and overcome gravitational pull is closest to the value we calculated, which is \( 8 \text{ kms}^{-1} \). Therefore, the correct option is (b) \( 8 \text{ kms}^{-1} \).

Key Concepts

Gravitational Pull

Gravitational pull is the force that attracts any object with mass towards the center of another object that also has mass. In the context of a satellite or a spaceship orbiting Earth, it is this force that keeps the object in orbit and prevents it from flying off into space. The strength of the gravitational pull depends on two main factors: the mass of the Earth and the distance of the object from Earth's center.

• The greater the mass of Earth, the stronger the gravitational pull it exerts.
• The closer the object to Earth's surface, the stronger the gravitational force acting on it.

Understanding gravitational pull is fundamental to calculating movements in space, as it dictates the velocity needed to enter and maintain an orbit.

Circular Orbit

A circular orbit is a path followed by an object around a larger body, like Earth, where the distance from the larger body remains constant. In this type of orbit, the gravitational force provides the necessary centripetal force to keep the object moving in a circle at a constant speed.
To maintain a circular orbit, the speed must be precise. If the speed is too slow, the object will fall due to gravity. If too fast, it could escape into another orbit or into space.
To calculate the speed required for this orbit, we use the formula for orbital velocity: \[ v_o = \sqrt{\frac{GM}{R}} \]

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

This speed ensures a balance, allowing the satellite to continuously "fall" around Earth without getting any closer to or farther away from it.

Satellite Motion

Satellite motion refers to the movement of a satellite as it orbits around a planet, like Earth. This motion is primarily governed by two forces: gravitational pull and the satellite's velocity. These forces ensure that the satellite maintains its path without deviations.

• Gravitational pull acts as the centripetal force, pulling the satellite towards Earth.
• Meanwhile, the satellite's velocity provides the centrifugal force needed to balance the gravitational pull, keeping it from crashing into Earth.

Satellites must maintain a specific velocity to stay in orbit. This velocity is termed as the **orbital velocity** and is different from the initial launch speed. Once the satellite reaches this velocity, the orbit becomes stable.

Escape Velocity

Escape velocity is the minimum speed necessary for an object to break free from the gravitational pull of a planet without further propulsion. It's the speed needed to overcome the gravitational force fully rather than balance it in orbit.
For Earth, this velocity is calculated using the formula:\[ v_e = \sqrt{\frac{2GM}{R}} \]

• \( G \) stands for the gravitational constant.
• \( M \) is the mass of the planet.
• \( R \) is the radius from the center of the Earth to the point of escape.

On Earth, this speed is approximately 11.2 km/s. This is higher than the orbital velocity because full escape requires overcoming Earth's gravity completely, unlike maintaining a balance as with orbital movement. Knowing the escape velocity is crucial for missions aiming to move beyond Earth's gravitational influence, such as traveling to other planets.