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Q33 PE

Expert-verifiedFound in: Page 191

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 $M=5.979\times {10}^{24}kg$****.**

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

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

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

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.

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

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

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

$M=5.979\times {10}^{24}kg$

This is the calculated mass of the earth.

The calculated value of mass is $M=5.979\times {10}^{24}kg$ 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.

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