Question

What is the force of gravity acting on a point of mass of $1\mathrm{kg}$ released from the height of $1\mathrm{m}$ from the surface of the earth; the radius of the earth is equal to $6.4\times {10}^6\text{m}$ ,mass of earth is equal to $6\times {10}^{24}\text{kg}$ ?

Answer 1

Step 1: Given value

Mass of earth, ${\mathrm{m}}_1=6\times {10}^{24}\mathrm{kg}$

Mass of the object ${\mathrm{m}}_2=1\mathrm{kg}$

The radius of the earth $\mathrm{r}=6.4\times {10}^6\mathrm{m}$

Universal gravitational constant $\mathrm{G}=6.67\times {10}^{-11}\frac{{\mathrm{Nm}}^2}{{\mathrm{kg}}^2}$

Step 2: To find

Determine the gravitational force acting on the particle.

Step 3: Finding the gravitational force acting on the particle

We know the law of gravitation:

$$\begin{gathered} \mathrm{F}=\left(\frac{\mathrm{G}{\mathrm{m}}_1{\mathrm{m}}_2}{{\mathrm{r}}^2}\right) \\ \Rightarrow =\frac{(6.67\times {10}^{-11}\times 6\times {10}^{24}\times 1)}{{(7.4\times {10}^6)}^2}\left(\because h=1\mathrm{m}\right) \\ \Rightarrow =\frac{(6.67\times {10}^{-11}\times 6\times {10}^{24}\times 1)}{(7.4\times 7.4\times {10}^{12})} \\ \Rightarrow =\frac{(40.02\times 10)}{(40.96)} \\ \Rightarrow =\frac{400.2}{54.76} \\ \Rightarrow =7.30\mathrm{N} \end{gathered}$$

Hence, the force acting on an object is $\mathrm{F}=7.30\mathrm{N}$.