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

Hint: 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{align}
  & {{\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{align} $
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{align}
  & \text{F}=53\cdot 36\times {{10}^{35}}\text{ N} \\
 & \text{F}=5\cdot 34\times {{10}^{36}}\text{ N} \\
\end{align} $ .

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}} $ .