Question

If masses of the two bodies are given as ${{\text{m}}_{1}}=2\times {{10}^{30}}Kg\text{ and }{{\text{m}}_{2}}=6\times {{10}^{24}}Kg$ separated by a distance of $\text{r}=1\cdot 5\times {{10}^{8}}m$ , what is the magnitude of gravitational force acting on them? $\left( \text{G}=6\cdot 67\times {{10}^{-11}}\text{ N}{{\text{m}}^{2}}\text{/k}{{\text{g}}^{2}} \right)$ .

Answer 1

Gravitational force is calculated by using the given formula:
${{\text{F}}_{\text{G}}}=\dfrac{\text{G}{{\text{m}}_{1}}{{\text{m}}_{2}}}{{{\text{r}}^{2}}}$
Where ${{\text{F}}_{\text{G}}}$ is the gravitational force acting on the body of mass ${{\text{m}}_{1}}$ due to ${{\text{m}}_{2}}$ which are separated by a distance of r.

Complete step by step solution
Here
$\begin{aligned} & {{\text{m}}_{1}}=2\times {{10}^{30}}\text{ kg} \\\ & {{\text{m}}_{2}}=6\times {{10}^{24}}\text{ kg} \\\ & \text{r}=1\cdot 5\times {{10}^{8}}\text{ m} \\\ & \text{G}=6\cdot 67\times {{10}^{-11}}\text{ N}{{\text{m}}^{2}}\text{ k}{{\text{g}}^{-2}} \\\ \end{aligned}$
As gravitational force is given by: ${{\text{F}}_{\text{G}}}=\dfrac{\text{G}{{\text{m}}_{1}}{{\text{m}}_{2}}}{{{\text{r}}^{2}}}$
So $\text{F}=6\cdot 67\times {{10}^{-11}}\times \dfrac{2\times {{10}^{30}}\times 6\times {{10}^{24}}}{1\cdot 5\times {{10}^{8}}}$
$\begin{aligned} & \text{F}=53\cdot 36\times {{10}^{35}}\text{ N} \\\ & \text{F}=5\cdot 34\times {{10}^{36}}\text{ N} \\\ \end{aligned}$ .

Note
The gravitational force being an attractive force has a direction also. Although the formula of calculation of force remains the same for two bodies i.e. force acting on ${{\text{m}}_{1}}$ due to ${{\text{m}}_{\text{2}}}$ is equal and opposite to the force acting on ${{\text{m}}_{2}}$ due to ${{\text{m}}_{1}}$ .