Chapter 18, Problem 112A

(a)

To determine

To compare: the force of gravity on a person due to the Sun at 1000 times farther away from its center and when the person was standing at its surface.

Explanation of Solution

Introduction:

Gravity is a force that forces a body with mass into the earth's core, or into some other physical object. Gravity plays an important part in the survival of earthly life.

Newton’s law of gravitation states that the force of gravity on a body of mass m from the Sun of mass Mcan be expressed as,

  $F_g=\frac{G{M}_{s}m}{r^2}$.................. (1)

Here, $G$ is the universal gravitational constant, $M_s$ is the mass of the sun, $m$ is the mass of a body, and $r$ is the distance from the centre of the sun.

The force of gravity on the person from the sun at his initial position can be written as,

  $F_g=\frac{G{M}_{s}m}{r_1{}^2}$.................. (2)

Here, $r_1$ is the initial position of the person.

Substitute $1000r_1$ for $r$ in equation (1),

  $\begin{array}{c}{F^{\prime }}_g=\frac{G{M}_{s}m}{{\left(1000r_1\right)}^2}\\ =\frac{G{M}_{s}m}{{10}^6{r}_1{}^2}\\ ={10}^{-6}\left(\frac{G{M}_{s}m}{r_1{}^2}\right)\\ {F^{\prime }}_g={10}^{-6}{F}_g\end{array}$

This is the force of gravity on the person from the sun at his new position,

Conclusion:

Hence, the force of gravity from the sun decreases by ${10}^{-6}$ times its initial value if the moves to a position which is 1000 times as far as he was initially from the sun.

(b)

To determine

To compare: the illuminance on his hand from the sun at the new position to what it was when he was standing on its surface.

Explanation of Solution

Introduction:

The illuminance on his hand by the sunlight is equal to the luminous flux of the sun divided by the area of the sphere whose radius is equal to his distance from the sun’s centre.

The expression for illuminance is given by the formula

  $E=\frac{P_{\text{s}}}{4\pi r^2}$..................(1)

Here, $P_s$ is the luminous flux of the sun and $r$ is the distance of the person the sun’s centre.

Use this relation to write the illuminance of his hand at his initial position,

  $E=\frac{P_{\text{s}}}{4\pi r_1{}^2}$..................(2)

Here, $r_1$ is the distance of the person from the sun initially.

Substitute $1000r_1$ for $r$ in equation (2), the illuminance of his hand by the sunlight at his new position,

  $\begin{array}{c}{E}^{\prime }=\frac{P_{\text{s}}}{4\pi {\left(1000r_1\right)}^2}\\ =\frac{P_{\text{s}}}{{10}^6\times 4\pi {\left(r_1\right)}^2}\\ ={10}^{-6}\times \left(\frac{P_{\text{s}}}{4\pi {\left(r_1\right)}^2}\right)\\ ={10}^{-6}\times E\\ \end{array}$

Conclusion:

Hence, the illuminance of his hand is decreased by ${10}^{-6}$ the time its initial value if the person moves to a new position which 1000 times as far as the person was initially from the sun.

(c)

To determine

To compare: the effect of distance upon the gravitational force and illuminance.

Explanation of Solution

Introduction:

The illuminance on his hand by the sunlight is equal to the luminous flux of the sun divided by the area of the sphere whose radius is equal to his distance from the sun’s centre. Gravity is the force by which earth or other body draws particles into its centre. The tidal force forces the planets in orbit around the earth.

From the above result of part (a) and (b), one can see that both the force of gravity and the illuminance decrease by the same amount with the increase in the distance from the Sun. in fact, both of them decreases as the square of the distance to Sun from the person. So, both forces of gravity and the illuminance follow the inverse square law.

Conclusion:

Hence, both the force of gravity and the illuminance follow the inverse square law.