Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 735

The escape velocity of a projectile from the earth is approximately (A) \(11.2 \mathrm{kms}^{-1}\) (B) \(112 \mathrm{kms}^{-1}\) (C) \(11.2 \mathrm{~ms}^{-1}\) (D) \(1120 \mathrm{kms}^{-1}\)

Short Answer

Expert verified
The escape velocity of a projectile from the earth is approximately (A) \(11.2 \mathrm{kms}^{-1}\).

Step by step solution

01

Understand the Escape Velocity Formula

The escape velocity formula is given by: \(v_e = \sqrt{\frac{2GM}{r}}\) where \(v_e\) = escape velocity \(G\) = gravitational constant, approximately \(6.674 \times 10^{-11} \frac{N(m/kg)^2}{m^2}\) \(M\) = mass of the earth, approximately \(5.972 \times 10^{24} kg\) \(r\) = radius of the earth, approximately \(6.371 \times 10^6 m\)
02

Substitute values and calculating the escape velocity

Plug the values of gravitational constant (G), mass of the Earth (M), and radius of the Earth (r) into the escape velocity formula and solve for the escape velocity (\(v_e\)). \(v_e = \sqrt{\frac{2 * (6.674 * 10^{-11}) * (5.972 * 10^{24})}{(6.371 * 10^6)}}\)
03

Perform the mathematical operations

Calculate the expression inside the square root. \(\frac{2 * (6.674 * 10^{-11}) * (5.972 * 10^{24})}{(6.371 * 10^6)} = 1.419 * 10^{17}\)
04

Find the square root

Now solve for \(v_e\). \(v_e = \sqrt{1.419 * 10^{17}} = 11.2 * 10^3 m/s\)
05

Interpret the result

The calculated escape velocity is \(11.2 * 10^3 \frac{m}{s}\), which can be expressed as \(11.2 \frac{km}{s}\). Therefore, the correct answer is (A) \(11.2 \mathrm{kms}^{-1}\).

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 of an object from the earth depends upon the mass of earth (M), its mean density ( \(p\) ), its radius (R) and gravitational constant (G), thus the formula for escape velocity is (A) \(U=\mathrm{R} \sqrt{[}(8 \pi / 3) \mathrm{Gp}]\) (C) \(\mathrm{U}=\sqrt{(2 \mathrm{GMR})}\) (D) \(U=\sqrt{\left[(2 \mathrm{GMR}) / \mathrm{R}^{2}\right]}\)

At what height over the earth's pole, the free fall acceleration decreases by one percent \(=\ldots \ldots \ldots \mathrm{km}(\mathrm{Re}=6400 \mathrm{~km})\) (A) 32 (B) 80 (C) \(1.253\) (D) 64

A satellite revolves around the earth in an elliptical orbit. Its speed (A) is the same at all points in the orbit (B) is greatest when it is closest to the earth (C) is greatest when it is farthest to the earth (D) goes on increasing or decreasing continuously depending upon the mass of the satellite

Two satellites \(\mathrm{A}\) and \(\mathrm{B}\) go round a planet in circular orbits having radii \(4 \mathrm{R}\) and \(\mathrm{R}\) respectively If the speed of satellite \(\mathrm{A}\) is \(3 \mathrm{v}\), then speed of satellite \(\mathrm{B}\) is (A) \((3 \mathrm{v} / 2)\) (B) \((4 \mathrm{v} / 2)\) (C) \(6 \mathrm{v}\) (D) \(12 \mathrm{v}\)

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

Essay Writing Skills

Read Explanation

Argumentative Essay

Read Explanation

Summary Text

Read Explanation

Semiotics

Read Explanation

Rhetoric

Read Explanation

Single Paragraph Essay

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