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Q33 PE

Expert-verified
Found in: Page 191

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# (a) Calculate Earth’s mass given the acceleration due to gravity at the North Pole is ${\text{9.830 m/s}}^{\text{2}}$ and the radius of the Earth is $6371km$ from pole to pole. (b) Compare this with the accepted value of $M=5.979×{10}^{24}kg$.

(a) The calculated mass of the earth is $M=5.979×{10}^{24}kg$

(b) The calculated value is matching with accepted value.

See the step by step solution

## Step 1: Definition of Gravity

Gravity is a universal phenomenon and is introduced by Newton and Derived the expression for gravitational force.

## Step 2: Calculating the mass of Earth

Earth’s radius ${\text{= R = 6371 x 10}}^{\text{3}}\text{m}$

Gravitational acceleration near the North Pole ${\text{= a = 9.83 m/s}}^{\text{2}}$

Earth’s mass $\text{= M = ?}$

The gravitational force is calculated as follows:

$\text{F = G}\frac{\text{Mm}}{{\text{R}}^{\text{2}}}$

Here, $m$ is object’s mass, $M$ is Earth’s mass, R Earth’s radius is the gravitation constant and F is the gravitational force. The force acting on the object is calculated as follows:$F=ma$ , here, is the object’s acceleration.

$\text{ma = G}\frac{\text{Mm}}{{\text{R}}^{\text{2}}}$

${\text{M = aR}}^{\text{2}}\text{/G}$

$M=\left(9.83×{\left(6371×{10}^{3}\right)}^{2}\right)/\left(6.637×{10}^{-11}\right)$

$M=5.979×{10}^{24}kg$

This is the calculated mass of the earth.

## Step 3: Determining difference between the calculated and accepted value

The calculated value of mass is $M=5.979×{10}^{24}kg$ which is exactly matching with the accepted value of the mass of the earth. The difference in percent between the computed and approved values is zero percent.