Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 725

A body of mass \(\mathrm{m} \mathrm{kg}\) starts falling from a point $2 \mathrm{R}\( above the earth's surface. Its \)\mathrm{K} . \mathrm{E}$. when it has fallen to a point ' \(\mathrm{R}\) ' above the Earth's surface $=\ldots \ldots \ldots \ldots J$ [R - Radius of Earth, M-mass of Earth G-Gravitational constant \(]\) (A) \((1 / 2)[(\mathrm{GMm}) / \mathrm{R}]\) (B) \((1 / 6)[(\mathrm{GMm}) / \mathrm{R}]\) (C) \((2 / 3)[(\mathrm{GMm}) / \mathrm{R}]\) (D) \((1 / 3)[(\mathrm{GMm}) / \mathrm{R}]\)

Short Answer

Expert verified
The short answer to the problem is: The kinetic energy (KE) of the object when it has fallen to a point R above the Earth's surface is given by option (B) \(\frac{1}{6}(\frac{GMm}{R})\).

Step by step solution

01

Identify the Initial and Final Points

We need to first identify the initial and final positions of the object. Initially, the object is at a height of 2R above the Earth's surface, and finally, it falls to a height of R above the Earth's surface.
02

Conservation of Mechanical Energy

According to the conservation of mechanical energy, the total mechanical energy (kinetic energy + gravitational potential energy) remains constant. In other words, the initial total mechanical energy equals the final total mechanical energy: Initial total mechanical energy = Final total mechanical energy
03

Calculate the Initial and Final Gravitational Potential Energies

We need to find the gravitational potential energy (GPE) at both initial and final positions. Since GPE is given by the formula: GPE = \(-\frac{GMm}{r}\) Initial GPE = \(-\frac{GMm}{3R}\) Final GPE = \(-\frac{GMm}{2R}\)
04

Calculate the Final Kinetic Energy (KE)

Since the body is initially at rest, its initial KE is equal to 0. Now we need to calculate the final kinetic energy when the fall reaches a height of R above the Earth's surface. We start by applying the conservation of mechanical energy: Initial KE + Initial GPE = Final KE + Final GPE 0 + \(-\frac{GMm}{3R}\) = Final KE + \(-\frac{GMm}{2R}\)
05

Solve for the Final Kinetic Energy (KE)

We can now solve for the final kinetic energy: Final KE = \(\frac{GMm}{2R} - \frac{GMm}{3R}\) Final KE = \(\frac{GMm}{6R}\) Thus, the correct answer is (B) \(\frac{1}{6}(\frac{GMm}{R})\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The escape velocity for a rocket from earth is \(11.2 \mathrm{kms}^{-1}\) value on a planet where acceleration due to gravity is double that on earth and diameter of the planet is twice that of earth will be $=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}$ (A) \(11.2\) (B) \(22.4\) (C) \(5.6\) (C) \(53.6\)

The earth revolves round the sun in one year. If distance between then becomes double the new period will be years. (A) \(0.5\) (B) \(2 \sqrt{2}\) (C) 4 (D) 8

Given mass of the moon is \((1718)\) of the mass of the earth and corresponding radius is \((1 / 4)\) of the earth, If escape velocity on the earth surface is \(11.2 \mathrm{kms}^{-1}\) the value of same on the surface of moon is $=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}$. (A) \(0.14\) (B) \(0.5\) (C) \(2.5\) (D) 5

The Gravitational P.E. of a body of mass \(\mathrm{m}\) at the earth's surface is \(-\mathrm{mgRe}\). Its gravitational potential energy at a height \(\operatorname{Re}\) from the earth's surface will be \(=\ldots \ldots \ldots\) here (Re is the radius of the earth) (A) \(-2 \mathrm{mgRe}\) (B) \(2 \mathrm{mgRe}\) (C) \((1 / 2) \mathrm{mg} \mathrm{Re}\) (D) \(-(1 / 2) \mathrm{mg} \operatorname{Re}\)

A small satellite is revolving near earth's surface. Its orbital velocity will be nearly \(=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}\). (A) 8 (B) 4 (C) 6 (D) \(11.2\)
See all solutions

Recommended explanations on English Textbooks

Textual Analysis

Read Explanation

Rhetorical Analysis Essay

Read Explanation

Rhetoric

Read Explanation

Linguistic Terms

Read Explanation

English Grammar

Read Explanation

Key Concepts in Language and Linguistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free