Chapter 5: Q32. (page 133)

A hypothetical planet has a radius 2.0 times that of Earth, but has the same mass. What is the acceleration due to gravity near its surface?

Expert verified

The acceleration due to gravity near the planet’s surface is $g_{p} = 2.45 m / s^{2}$ .

Step by step solution

The radius of the planet is $r_{p} = 2 r_{e}$ .

The radius of the Earth is $r_{e}$ .

In this problem, the relation of acceleration due to gravity at any location on the planet’s surface will be utilized.

The relation of gravitational acceleration is given by,

$g_{p} = \frac{G m_{p}}{{r_{p}}^{2}}$

Here, $m_{p}$ is the mass of the planet equal to the mass of the Earth and G is the gravitational constant.

On plugging the values in the above relation.

$\begin{matrix} g_{p} = \frac{G m_{e}}{{(2 r_{e})}^{2}} \\ g_{p} = \frac{1}{4} \frac{G m_{e}}{{r_{e}}^{2}} \\ g_{p} = \frac{1}{4} g_{e} \end{matrix}$

Here, $g_{e}$ is the gravitational acceleration of the Earth.

Substitute the values in the above equation.

$\begin{array}{l} g_{p} = \frac{1}{4} (9.80 m / s^{2}) \\ g_{p} = 2.45 m / s^{2} \end{array}$

Thus, $g_{p} = 2.45 m / s^{2}$ is the required acceleration due to gravity.

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