Question

Masses M and m attract each other with a gravitational force of magnitude F. Mass m is replaced with a mass 3 cm, and it is moved four times farther away. Now, what is the magnitude of the force?

Short answer

The magnititude of the force is$F=1.25\times {10}^{-11}\times \left(\frac{mM}{d^2}\right)\mathrm{N}$.

Step-by-step solution

identification of the given data

- Mass$m=3m$

-$d^{\text{'}}=4d$

Gravitational Force

Bodies that have mass and are at a certain distance attract each other. This attraction is the gravitational force.

Calculation of the force magnitude

By Newton's law, the magnitude of the force is written as:

$F=G\frac{m_1\times m_2}{d^2}$

Here, $F$- Magnitude of gravitational force,$m_1$- the mass of object $1,m_2$- the mass of object 2, $\mathrm{d}$- Distance separating two objects, and $\mathrm{G}$- Universal gravitation constant $\left(6.67\times {10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\right)$

$F=G\frac{m_1\times m_2}{d^{\text{'}2}}$

$F=\left(6.67\times {10}^{-11}\right)\frac{M\times 3m}{(4d{)}^2}$

$F=1.25\times {10}^{-11}\times \left(\frac{mM}{d^2}\right)\mathrm{N}$

Thus, the value of force is$F=1.25\times {10}^{-11}\times \left(\frac{mM}{d^2}\right)\mathrm{N}$.