# (a) Calculate the magnitude of the gravitational force exerted on a 432-kg satellite that is a...

## Question:

(a) Calculate the magnitude of the gravitational force exerted on a 432-kg satellite that is a distance of 2.55 times earth radii from the centre of the earth.

(b) What is the magnitude of the gravitational force exerted on the earth by the satellite?

(c) Determine the magnitude of the satellite's acceleration.

(d) What is the magnitude of the earth's acceleration?

## Gravitation Force:

The attraction force that every object and energy exert on each other is called gravitational force. This force is very weak and not considered for the analysis of force for smaller bodies. But it plays an important role in the analysis of big bodies (like planets and stars).

## Answer and Explanation: 1

**Given data**

- The mass of satellite; {eq}m = 432\;{\rm{kg}} {/eq}

- The distance of the satellite from the earth is 2.55 times of radius of the Earth; {eq}r = 2.55 \times {R_{earth}} {/eq}

From the physical parameters of Earth, the radius of the Earth is

{eq}{R_{earth}} = 6371\;{\rm{km}} {/eq}

And the mass of the Earth is,

{eq}M = 5.97 \times {10^{24}}\;{\rm{kg}}{/eq}

**(a)**

Calculate the gravitational force exerted on satellite by the Earth,

{eq}\begin{align*} {F_1} &= \dfrac{{GmM}}{{{r^2}}}\\ {F_1} &= \dfrac{{GmM}}{{{{\left( {2.55 \times {R_{earth}}} \right)}^2}}}\\ &= \dfrac{{6.67 \times {{10}^{ - 11}}\;{\rm{N}} \cdot {{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2} \times 432\;{\rm{kg}} \times 5.97 \times {{10}^{24}}\;{\rm{kg}}}}{{{{\left( {2.55 \times 6371 \times {{10}^3}\;{\rm{m}}} \right)}^2}}}\\ &= 651.8\;{\rm{N}} \end{align*} {/eq}

Thus, the gravitational force exerted on the satellite by the Earth is **651.8 N**.

**(b)**

Calculate the gravitational force exerted on the Earth by the satellite,

{eq}\begin{align*} {F_2} &= \dfrac{{GmM}}{{{r^2}}}\\ {F_2} &= \dfrac{{GmM}}{{{{\left( {2.55 \times {R_{earth}}} \right)}^2}}}\\ &= \dfrac{{6.67 \times {{10}^{ - 11}}\;{\rm{N}} \cdot {{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2} \times 432\;{\rm{kg}} \times 5.97 \times {{10}^{24}}\;{\rm{kg}}}}{{{{\left( {2.55 \times 6371 \times {{10}^3}\;{\rm{m}}} \right)}^2}}}\\ &= 651.8\;{\rm{N}} \end{align*} {/eq}

Thus, the gravitational force exerted on the Earth by the satellite is **651.8 N**.

**(c)**

Apply the second law of newton to calculate the acceleration of the satellite,

{eq}\begin{align*} {F_1} &= m{a_{sat}}\\ {a_{sat}} &= \dfrac{{{F_1}}}{m}\\ &= \dfrac{{651.8\;{\rm{N}}}}{{432\;{\rm{kg}}}}\\ &= 1.5\;{\rm{m}}/{{\rm{s}}^2} \end{align*} {/eq}

Thus, the acceleration of the satellite is {eq}1.5\;{\rm{m}}/{{\rm{s}}^2}{/eq}.

**(d)**

Apply the second law of newton to calculate the acceleration of the Earth,

{eq}\begin{align*} {F_2} &= M{a_{earth}}\\ {a_{earth}} &= \dfrac{{{F_2}}}{M}\\ &= \dfrac{{651.8\;{\rm{N}}}}{{5.97 \times {{10}^{24}}\;{\rm{kg}}}}\\ &= 1.09 \times {10^{ - 22}}\;{\rm{m}}/{{\rm{s}}^2} \end{align*} {/eq}

Thus, the acceleration of the Earth is {eq}1.09 \times {10^{ - 22}}\;{\rm{m}}/{{\rm{s}}^2}{/eq}.

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Chapter 15 / Lesson 17Earth's gravitational pull is often misunderstood, but without it, life on Earth would be impossible. In this lesson, we'll define the gravitational pull and give some examples of how it is used. A quiz is provided to test your understanding.