/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 If \(M\) is mass of the earth an... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 61

If \(M\) is mass of the earth and \(R\) its radius, the ratio of the gravitational acceleration and the gravitational constant is (A) \(\frac{R^{2}}{M}\) (B) \(\frac{M}{R^{2}}\) (C) \(M R^{2}\) (D) \(\frac{M}{R}\)

Short Answer

Expert verified
The ratio of the gravitational acceleration (g) to the gravitational constant (G) is \(\frac{M}{R^{2}}\), which corresponds to option (B).

Step by step solution

01

Write the formula for gravitational acceleration.

The formula for gravitational acceleration (g) near the surface of the Earth can be derived from Newton's law of universal gravitation, which is given by: \(g = \frac{G M}{R^{2}}\) where - g is the gravitational acceleration - G is the gravitational constant - M is the mass of the Earth - R is the radius of the Earth
02

Divide gravitational acceleration by gravitational constant.

Now we need to find the ratio of the gravitational acceleration (g) to the gravitational constant (G). To do this, divide g by G: \(\frac{g}{G} = \frac{\frac{G M}{R^{2}}}{G}\)
03

Simplify the expression.

Next, simplify the expression by canceling out G from both the numerator and the denominator: \(\frac{g}{G} = \frac{M}{R^{2}}\)
04

Compare the simplified expression with the given options.

Now, let's compare our simplified expression with the options given in the problem: (A) \(\frac{R^{2}}{M}\) (B) \(\frac{M}{R^{2}}\) (C) \(M R^{2}\) (D) \(\frac{M}{R}\) As we can see, option (B) matches our simplified expression.
05

Choose the correct answer.

Based on the comparison in Step 4, the correct answer is: (B) \(\frac{M}{R^{2}}\)

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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 least count of a stop watch is \(1 / 5 \mathrm{~s}\). The time of 20 oscillations of a pendulum is measured to be \(25 \mathrm{~s}\). The minimum percentage error in the measurement of time will be (A) \(0.1 \%\) (B) \(0.8 \%\) (C) \(1.8 \%\) (D) \(8 \%\)

The vectors \(\vec{A}=3 \hat{i}-2 \hat{j}+\hat{k}, \vec{B}=\hat{i}-3 \hat{j}+5 \hat{k}\) and \(\vec{C}=\) \(2 \hat{i}+\hat{j}-4 \hat{k}\) form a triangle, then the triangle is (A) right angled triangle (B) lateral triangle (C) isolated triangle (D) None of these

Find a vector \(\vec{x}\) which is perpendicular to both \(\vec{A}\) and \(\vec{B}\) but has magnitude equal to that of \(\vec{B}\). Rule: Inter change coeff. of \(\hat{i}\) and \(\hat{j}\) and change sign of one of the vectors. \(\vec{A}=3 \hat{i}-2 \hat{j}+\hat{k}, \vec{B}=4 \hat{i}+3 \hat{j}-2 \hat{k}\) (A) \(\frac{1}{\sqrt{10}}(\hat{i}+10 \hat{j}+17 \hat{k})\) (B) \(\frac{1}{\sqrt{10}}(\hat{i}-10 \hat{j}+17 \hat{k})\) (C) \(\sqrt{\frac{29}{390}}(\hat{i}-10 \hat{j}+17 \hat{k})\) (D) \(\sqrt{\frac{29}{390}}(\hat{i}+10 \hat{j}+17 \hat{k})\)

Let \(\left[\epsilon_{0}\right]\) denote the dimensional formula of the permittivity of vacuum. If \(M=\) mass, \(L=\) length, \(T=\) time and \(A=\) electric current, then (A) \(\left[\epsilon_{0}\right]=\left[M^{-1} L^{-3} T^{4} A^{2}\right]\) (B) \(\left[\epsilon_{0}\right]=\left[M^{-1} L^{2} T^{-1} A^{-2}\right]\) (C) \(\left[\in_{0}\right]=\left[\begin{array}{ll} \left.M^{-1} L^{2} T^{-1} A\right]\end{array}\right.\) (D) \(\left[\epsilon_{0}\right]=\left[M^{-1} L^{-3} T^{2} A\right]\)

If \(|\vec{A} \times \vec{B}|=\sqrt{3} \vec{A} \cdot \vec{B}\), then the value of \(|\vec{A}+\vec{B}|\) is (A) \(\left(A^{2}+B^{2}+A B\right)^{1 / 2}\) (B) \(\left(A^{2}+B^{2}+\frac{A B}{\sqrt{3}}\right)^{1 / 2}\) (C) \((A+B)\) (D) \(\left(A^{2}+B^{2}+\sqrt{3} A B\right)^{1 / 2}\)
See all solutions

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Read Explanation

Thermodynamics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Linear Momentum

Read Explanation

Quantum Physics

Read Explanation

Energy Physics

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.