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Chapter 13: Q1P (page 378)

A massMis split into two parts, $m$ and role="math" localid="1657189921712" $M - m$ , which are0 then separated by a certain distance. What ratio $m - M$ maximizes the magnitude of the gravitational force between the parts?

Short Answer

Expert verified

The magnitude of the gravitational force between the parts is maximized at $\frac{1}{2}$

Step by step solution

01

The given data

A mass Mis split into two parts: m and $M - m$

02

Understanding the concept of Newton’s law of gravitation

The problem is based on the concept of Newton鈥檚 law of Gravitation. If the magnitude of the force is to be maximum, its derivative to massmwill be zero. Hence this law is used to find the ratiom / M.

Formula:

Gravitational Force, $F = \frac{G M m}{r^{2}}$ (i)

03

Calculating the value of m / M that maximizes the force

Using equation (i), the gravitational force between the two parts m and $M - m$ of the mass M can be given as:

$F = \frac{G 脳 m 脳 (M - m)}{r^{2}} = \frac{G}{r^{2}} 脳 (m M - m^{2})$

Differentiating with respect m:

$\frac{d f}{d m} = \frac{G}{r^{2}} 脳 (M - 2 m)$

For the force to be maximum, we should have,

$\frac{d f}{d m} = 0$

$\frac{G}{r^{2}} 脳 (M - 2 m) = 0$

Hence,

$M - 2 m = 0 M = 2 m$

Therefore, the gravitational force is maximum at $\frac{1}{2}$

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Most popular questions from this chapter

Figure 13-43 gives the potential energy functionU(r) of aprojectile, plotted outward from the surface of a planet of radius $R_{s}$ . If the projectile is launched radially outward from the surfacewith a mechanical energy of $- 2.0 脳 10^{9} J$ , what are (a) its kineticenergy at radius $r = 1.25 R_{s}$ and (b) itsturning point (see Module 8-3)in terms of $R_{s}$ ?

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