Chapter 13, Problem 33A

What is the magnitude and direction of the gravitational force that acts on a woman who weighs 500 N at the surface of Earth?

To determine

The magnitude and direction of the gravitational force acting on the woman.

Answer to Problem 33A

The gravitational force acting on the woman is $500\text{N}\left(\begin{array}{l}\text{Acts towards the }\\ \text{centre of the Earth}\end{array}\right)$ .

Explanation of Solution

Given:

The weight of the woman is $W=500\text{N}$ .

Formula used:

Consider a person of mass $m$ on the surface of the Earth.

The expression for the acceleration due to gravity $\left(\overrightarrow{g}\right)$ is,

  $\overrightarrow{g}=G\frac{M}{r^2}$

Here, $M$ is the mass of the Earth and $G$ is the gravitational constant.

The expression for the force of gravity $\left(F\right)$ acting on the object is,

  $F=G\frac{Mm}{r^2}$

Here, $M$ is the mass of the Earth and $G$ is the gravitational constant.

The gravitational force on a person acts towards the centre of the Earth.

Calculation:

The force of gravity $\left(F\right)$ acting on the woman is

  $\begin{array}{l}F=G\frac{Mm}{r^2}\\ F=m\left(G\frac{M}{r^2}\right)\\ F=mg\\ F=\text{Weight}\text{of}\text{the}\text{woman}\\ F=500\text{N}\left(\begin{array}{l}\text{Acts towards the }\\ \text{centre of the Earth}\end{array}\right)\end{array}$

Conclusion:

Therefore, the force of gravity acting on the woman is $500\text{N}\left(\begin{array}{l}\text{Acts towards the }\\ \text{centre of the Earth}\end{array}\right)$ .