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Chapter 9: Problem 51

The escape velocity on the surface of the earth is \(11.2\) \(\mathrm{km} / \mathrm{sec}\). If mass and radius of a planet are 4 and 2 times respectively than that of earth. The escape velocity from the planet will be: (a) \(11.2 \mathrm{~km} / \mathrm{sec}\) (b) \(1.112 \mathrm{~km} / \mathrm{sec}\) (c) \(15.8 \mathrm{~km} / \mathrm{sec}\) (d) \(22.4 \mathrm{~km} / \mathrm{sec}\)

Short Answer

Expert verified
(c) 15.8 \(\mathrm{km/s}\)

Step by step solution

01

Understanding Escape Velocity Formula

The formula for escape velocity \( v_e \) is given by \( v_e = \sqrt{\frac{2GM}{R}} \), where \( G \) is the gravitational constant, \( M \) is the mass of the celestial body, and \( R \) is its radius.
02

Substituting Earth's Values

For Earth, the escape velocity \( v_{e, \text{earth}} = 11.2 \text{ km/s} \). This means \( 11.2 = \sqrt{\frac{2G M_{\text{earth}}}{R_{\text{earth}}}} \).
03

Determine Planet's Parameters

The planet's mass is 4 times Earth's mass: \( M_{\text{planet}} = 4M_{\text{earth}} \). The planet's radius is 2 times Earth's radius: \( R_{\text{planet}} = 2R_{\text{earth}} \).
04

Substituting Planet's Values into Escape Velocity Formula

Using the formula \( v_{e, \text{planet}} = \sqrt{\frac{2G \times 4M_{\text{earth}}}{2R_{\text{earth}}}} \), simplify the fraction: \( v_{e, \text{planet}} = \sqrt{\frac{4 \times 2G M_{\text{earth}}}{2R_{\text{earth}}}} = \sqrt{2 \times \frac{2GM_{\text{earth}}}{R_{\text{earth}}}} \).
05

Applying Earth's Escape Velocity

Since \( v_e = \sqrt{\frac{2GM_{\text{earth}}}{R_{\text{earth}}}} = 11.2 \text{ km/s} \) for Earth, the escape velocity for the new planet becomes \( v_{e, \text{planet}} = \sqrt{2} \times 11.2 \).
06

Calculating Final Escape Velocity

The value of \( \sqrt{2} \approx 1.414 \). Therefore, the escape velocity for the planet is \( v_{e, \text{planet}} = 1.414 \times 11.2 \approx 15.8 \text{ km/s} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Constant
The gravitational constant, denoted as \( G \), is a fundamental constant that plays a key role in the calculation of gravitational forces between masses. It is a crucial component of Newton's law of universal gravitation, encapsulating the strength of the gravitational interaction.

In the context of escape velocity, \( G \) allows us to determine the speed needed for an object to break free from the gravitational pull of a celestial body. Understanding this constant helps us comprehend how different mass distributions and distances affect gravitational attraction.
  • Gravitational constant value: \( G = 6.674 \times 10^{-11} \, \mathrm{m^3 \cdot kg^{-1} \cdot s^{-2}} \).
  • It is a universal constant, meaning it applies everywhere in the universe.
  • Facilitates calculation of gravitational force: \( F = \frac{G M_1 M_2}{r^2} \), where \( M_1 \) and \( M_2 \) are masses and \( r \) is the distance between their centers.
Mass and Radius Relationship
The relationship between mass and radius is crucial for determining escape velocity, which is the speed needed to leave a planet's gravitational field without further propulsion.

This relationship helps in calculating how the mass and radius of a celestial body affects the escape velocity from its surface. If a planet is more massive or larger than Earth, it affects how much speed is necessary to overcome its gravity.
  • Escape velocity \( v_e = \sqrt{\frac{2 G M}{R}} \) depends on mass \( (M) \) and radius \( (R) \).
  • Increasing a planet's mass while keeping radius the same increases escape velocity, as more energy is needed to overcome stronger gravity.
  • Increasing a planet's radius while keeping mass the same decreases escape velocity, as the gravitational influence diminishes with distance.
Celestial Mechanics
Celestial mechanics is the branch of astronomy that deals with the motions and gravitational forces of celestial objects. It鈥檚 the study that allows us to understand how planets, moons, and stars move through space, governed by gravitational laws.

This discipline plays a significant role in calculating escape velocity, as it considers the mass and radius of celestial bodies. Knowledge gained through celestial mechanics helps in space exploration, satellite deployment, and understanding orbital dynamics.
  • Involves the use of mathematical models and physics principles.
  • Important for mission planning in space travel and understanding planetary bodies.
  • Helps in determining stable orbits and potential energy requirements for travel.
AIEEE Physics Problems
The All India Engineering Entrance Examination (AIEEE) often includes physics problems like the calculation of escape velocity. Solving these problems requires understanding and applying fundamental physics concepts such as those involving the gravitational constant and celestial mechanics.

These problems not only test knowledge but also problem-solving skills and the ability to apply theoretical concepts to practical scenarios.
  • Focus on real-world applications of physics principles.
  • Help prepare students for engineering studies and careers.
  • Bridge the gap between textbook learning and practical applications.

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Most popular questions from this chapter

At the surface of a certain planet acceleration due to gravity is one-quarter of that on the earth. If a brass ball is transported to this planet, then which one of the following statements is not correct? (a) The mass of the brass ball on this planet is quarter of its mass as measured on the earth (b) The brass ball has same mass on the other planet as on the earth (c) The brass ball has same volume on the other planet as on the earth (d) None of the above

Two satellites \(S\) and \(S^{\prime}\) revolve around the earth at distances \(3 R\) and \(6 R\) from the centre of earth. Their periods of revolution will be in the ratio: (a) \(1: 2\) (b) \(2: 1\) (c) \(1: 2^{1.5}\) (d) \(1: 2^{0.67}\)

The escape velocity of a body on the surface of the earth is \(11.2 \mathrm{~km} / \mathrm{sec}\). If the earth's mass increases to twice its present value and radius of the earth becomes half, the escape velocity becomes: (a) \(5.6 \mathrm{~km} / \mathrm{sec}\) (b) \(11.2 \mathrm{~km} / \mathrm{sec}\) (c) \(22.4 \mathrm{~km} / \mathrm{sec}\) (d) \(44.8 \mathrm{~km} / \mathrm{sec}\)

The period of a satellite in a circular orbit of radius \(R\) is \(T\). What is the period of another satellite in a circular orbit of radius \(4 R\) ? (a) \(4 T\) (b) \(\frac{T}{8}\) (c) \(\frac{T}{4}\) (d) \(8 T\)

A projectile is fired vertically upwards from the surface of the earth with a velocity \(k v_{e}\), where \(v_{e}\) is the escape velocity and \(k<1\). If \(R\) is the radius of the earth, the maximum height to which it will rise measured from the centre of earth will be : (neglect air resistance) (a) \(\frac{1-k^{2}}{R}\) (b) \(\frac{R}{1-k^{2}}\) (c) \(R\left(1-k^{2}\right)\) (d) \(\frac{R}{1+k^{2}}\)
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