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Chapter 6: Problem 673

Correct form of gravitational law is (A) \(\mathrm{F}=-\left[\left(\mathrm{Gm}_{1} \mathrm{~m}_{2}\right) / \mathrm{r}^{2}\right]\) (B) \(\mathrm{F}^{-}=-\left[\left(\mathrm{Gm}_{1} \mathrm{~m}_{2}\right) / \mathrm{r}^{2}\right]\) (C) \(\mathrm{F}^{-}=-\left[\left(\mathrm{Gm}_{1} \mathrm{~m}_{2}\right) / \mathrm{r}^{2}\right] \hat{\mathrm{r}}\) (B) \(\mathrm{F}^{\rightarrow}=-\left[\left(\mathrm{Gm}_{1} \mathrm{~m}_{2}\right) / \mathrm{r}^{3}\right] \mathrm{r}^{-}\)

Short Answer

Expert verified
The correct form of the gravitational law is given by Option C: \(F^{-} = -\frac{(Gm_1m_2)}{r^2}\hat{r}\), as it includes the gravitational constant, the masses of both objects, the distance between their centers, and the force vector \(\hat{r}\) pointing along the line connecting the centers of the objects.

Step by step solution

01

Option A

Option A states: \(F = -\frac{(Gm_1m_2)}{r^2}\) This option does include the gravitational constant and the masses of both objects, and the force goes inversely with the square of the distance between them. However, this option does not include any direction information.
02

Option B

Option B states: \(F^{-} = -\frac{(Gm_1m_2)}{r^2}\) This option is nearly the same as option A, but with a minus sign. The inclusion of a minus sign could indicate the attractive nature of the gravitational force, but this option still does not include any direction information.
03

Option C

Option C states: \(F^{-} = -\frac{(Gm_1m_2)}{r^2}\hat{r}\) This option includes the gravitational constant, the masses of both objects, the distance between their centers, and a unit vector in the radial direction, \(\hat{r}\), which indicates the direction of the force. This seems to be the correct form of the gravitational law.
04

Option D

Option D states: \(F^{\rightarrow} = -\frac{(Gm_1m_2)}{r^3}r^{-} \) This option includes the gravitational constant and the masses of both objects. However, the force is inversely proportional to the cube of the distance, not the square, which is incorrect. Additionally, this option does not have a clear direction. Based on the analysis of all the options, the correct form of the gravitational law is found in Option C.

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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, often symbolized as \(G\), is a fundamental part of Newton's Law of Universal Gravitation. This constant is crucial because it sets the scale for gravitational force between two masses. - The standard value of the gravitational constant is approximately \(6.674 \times 10^{-11} \text{Nm}^2/ ext{kg}^2\). - It essentially quantifies the strength of gravity and how masses interact over distances.In the context of the gravitational law, \(G\) ensures that the force calculated is accurate. This is why it appears in every term of the gravitational law equations listed in the exercise options. Understanding \(G\) is key to solving gravity-related problems involving planets, stars, or even everyday objects.
Inverse Square Law
Newton's Law of Universal Gravitation is governed by the inverse square law, which implies that the gravitational force between two objects decreases with the square of the distance between them. - Mathematically, this relationship is expressed as \(F \propto \frac{1}{r^2}\), highlighting that as distance \(r\) increases, the force \(F\) significantly decreases.- This principle is why planets orbit the Sun in a stable path and explains why gravity is much weaker at greater distances.Understanding the inverse square law is essential for recognizing why option D in the exercise is incorrect. Option D attempts to suggest an inverse cube relationship, which conflicts with the true nature of gravitational interactions.
Vector Notation
Vector notation is indispensable for accurately describing forces, including gravitational forces, because it includes both magnitude and direction. - In physics, forces are vectors, meaning they have a size and a direction, depicted by arrows and often denoted with vectors like \(\mathbf{F}\).- In the gravitational law, the unit vector \(\hat{r}\) is used to indicate the direction of the gravitational force, pointing from one mass to another.Option C in the exercise introduces \(\hat{r}\), the unit vector, making it the correct choice as it reflects both the magnitude and direction of the gravitational force. Understanding vector notation is crucial to fully grasping how gravitational forces operate in real-world scenarios.

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Most popular questions from this chapter

A particle of mass \(10 \mathrm{~g}\) is kept on the surface of a uniform sphere of mass \(100 \mathrm{~kg}\) and radius \(10 \mathrm{~cm}\). Find the work to be done against the gravitational force between them to take the particle is away from the sphere \(\left(\mathrm{G}=6.67 \times 10^{-11} \mathrm{SI}\right.\) unit \()\) (A) \(6.67 \times 10^{-9} \mathrm{~J}\) (B) \(6.67 \times 10^{-10} \mathrm{~J}\) (C) \(13.34 \times 10^{-10} \mathrm{~J}\) (D) \(3.33 \times 10^{-10} \mathrm{~J}\)

The mass of a space ship is \(1000 \mathrm{~kg} .\) It is to be launched from earth's surface out into free space the value of \(\mathrm{g}\) and \(\mathrm{R}\) (radius of earth) are \(10 \mathrm{~ms}^{-2}\) and \(6400 \mathrm{~km}\) respectively the required energy for this work will be \(=\ldots \ldots \ldots .\) J (A) \(6.4 \times 10^{11}\) (B) \(6.4 \times 10^{8}\) (C) \(6.4 \times 10^{9}\) (D) \(6.4 \times 10^{10}\)

The escape velocity for a body projected vertically upwards from the surface of earth is \(11 \mathrm{kms}^{-1}\). If the body is projected at an angle of \(45^{\circ}\) with the vertical, the escape velocity will be \(\ldots \ldots \ldots \mathrm{kms}^{-1}\) (A) \((11 / \sqrt{2})\) (B) \(11 \sqrt{2}\) (C) 22 (D) 11

let \(\mathrm{V}\) and \(\mathrm{E}\) denote the gravitational potential and gravitational field at a point. Then the match the following \(\begin{array}{ll}\text { Table }-1 & \text { Table }-2\end{array}\) (A) \(\mathrm{E}=0, \mathrm{~V}=0\) (P) At center of spherical shell (B) \(\mathrm{E} \neq 0, \mathrm{~V}=0\) (Q) At center of solid sphere (C) \(\mathrm{V} \neq 0, \mathrm{E}=0\) (R) at centre of circular ring (D) \(\mathrm{V} \neq 0, \mathrm{E} \neq 0\) (S) At centre of two point masses of equal magnitude (T) None

Direction (Read the following questions and choose) (A) If both Assertion and Reason are true and the Reason is correct explanation of assertion (B) If both Assertion and Reason are true, but reason is not correct explanation of the Assertion (C) If Assertion is true, but the Reason is false (D) If Assertion is false, but the Reason is true Assertion: The value of acc. due to gravity \((\mathrm{g})\) does not depend upon mass of the body Reason: This follows from \(\mathrm{g}=\left[(\mathrm{GM}) / \mathrm{R}^{2}\right]\), where \(\mathrm{M}\) is mass of planet (earth) and \(\mathrm{R}\) is radius of planet (earth) (a) \(\mathrm{A}\) (b) \(\mathrm{B}\) (c) \(\mathrm{C}\) (d) D
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