Chapter 13, Problem 56A

By what factor would your weight change if the Earth’s diameter were doubled and its mass were also doubled?

To determine

The changing factor of weight of the person for the given changes.

Answer to Problem 56A

The weight of the person changes by a factor $\left(\frac{1}{2}\right)$ , when the Earth’s diameter is doubled and the mass of the Earth is doubled.

Explanation of Solution

Given:

Consider the weight of the person is denoted by $W$ .

Diameter of Earth is doubled and mass is also doubled.

Formula used:

The Newton’s equation for gravitational force is,

  $F_{\text{g}}=G\frac{mM}{R}$

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

Calculation:

The weight of the person is equal to the gravitational force acting on the person. The initial weight of the person is,

  $\begin{array}{l}{W}_{\text{initial}}=F_{\text{g}}\\ W_{\text{initial}}=G\frac{mM}{R}\\ W_{\text{initial}}=G\frac{mM}{\left(\frac{D}{2}\right)}\\ W_{\text{initial}}=4G\frac{mM}{D^2}\end{array}$

Here, $D$ is the diameter of the Earth.

Consider the weight of the person when the mass $\left(M\right)$ of the Earth is doubled and the diameter $\left(D\right)$ of the Earth is doubled, is $W_{\text{modified}}$ .

The modified weight of the person is,

  $\begin{array}{l}{W}_{\text{modified}}=4G\frac{m\left(2M\right)}{{\left(2D\right)}^2}\\ W_{\text{modified}}=4G\left(\frac{2mM}{4D^2}\right)\\ W_{\text{modified}}=\frac{1}{2}\left(4G\frac{mM}{D^2}\right)\\ W_{\text{modified}}=\frac{1}{2}{W}_{\text{initial}}\end{array}$

Conclusion:

Thus, the weight of the person when the mass of the Earth is doubled and the diameter of the Earth is doubled, is half the weight of the same person in the initial case.