Question

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

Short answer

The magnitude 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, m₁- the mass of object 1, m₂- the mass of object 2, d – Distance separating two objects, and G – Universal gravitation constant ( $6.67\times {10}^{-11}N\cdot m^2/k{g}^2$).

role="math" $$\begin{gathered} F=G\frac{m_1\times m_2}{d^{\text{'}2}} \\ F=\left(6.67\times {10}^{-11}\right)\frac{M\times 3m}{(4d{)}^2} \\ F=1.25\times {10}^{-11}\times \left(\frac{mM}{d^2}\right)\mathrm{N} \end{gathered}$$

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