Chapter 5: Q37. (page 134)

A hypothetical planet has a mass 2.80 times that of Earth, but has the same radius. What is g near its surface?

Short Answer

Expert verified

The gravitational acceleration of the planet is $27.4 m / s^{2}$ .

Step by step solution

Step 1. Given Data

The mass of the planet is $m_{p} = 2.80 m_{e}$ .

Step 2. Understanding the acceleration due to gravity

In this problem, the equation of acceleration due to gravity at any location will be applied. Also, the value of mass as well as radius of the Earth will be used.

Step 3. Estimating the acceleration due to gravity on the planet

The relation of acceleration due to gravityis given by,

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

Here, G is the gravitational constant and $r$ is the radius of the planet whose value is same as radius of the Earth.

On plugging the values in the above relation.

$\begin{array}{l} g_{p} = G (\frac{2.80 m_{e}}{r^{2}}) \\ g_{p} = 2.80 (g) \\ g_{p} = 2.80 (9.80 m / s^{2}) \\ g_{p} = 27.4 m / s^{2} \end{array}$

Thus, $g_{p} = 27.4 m / s^{2}$ is the required acceleration due to gravity.

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A hypothetical planet has a mass 2.80 times that of Earth, but has the same radius. What is g near its surface?

Short Answer

Step by step solution

Step 1. Given Data

Step 2. Understanding the acceleration due to gravity

Step 3. Estimating the acceleration due to gravity on the planet

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