Question

Two objects attract each other gravitationally with a force of $\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{N}$when they are 0.25 m apart. Their total mass is 4.00 kg. Find their individual masses.

Short answer

The masses of two objects are 0.06 kg and 3.94 kg.

Step-by-step solution

Step 1. Understanding the Newton’s law of universal gravitation

Newton’s law of universal gravitation describes the force of attraction between two objects establishes an inverse relation with the square of the distance between them. It varies directly with the product of the masses of the objects.

If $m_1$ and $m_2$are the masses of the objects, and r is the distance between the objects, then the expression for the gravitational force is given by:

$F_{\mathrm{G}}=\frac{G{m}_1{m}_2}{r^2}$ ...(i)

Here, $G$is the universal gravitational constant.

Step 2. Identification of the given information

• The total mass of the objects is, $m_1+m_2=4.00\mathrm{kg}$.
• The magnitude of the gravitational force between the objects is, $F_{\mathrm{G}}=2.5\times {10}^{-10}\mathrm{N}$.
• The distance between the objects is, $r=0.25\mathrm{m}$.

Step 3. Evaluation of the individual masses of the objects

On substituting given values in (i), the gravitational force between the objects can be written as:

$F_{\mathrm{G}}=\frac{G{m}_1{m}_2}{r^2}$

On putting the given values, you will get:

$\begin{array}{c}2.5\times {10}^{-10}\mathrm{N}=\frac{\left(6.67\times {10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\right)\times m_1(4.00-m_1)}{{(0.25\mathrm{m})}^2}\\ 4.00m_1-m_1^2=0.234\\ m_1^2-4.00m_1+0.234=0\end{array}$

The above equation is quadratic. On solving this equation, you get two values of $m_1$.

$m_1=3.94\mathrm{or}0.06$

If $m_1=3.94\mathrm{kg}$, then the value of $m_2=\left(4.00-3.94\right)\mathrm{kg}=0.06\mathrm{kg}$.

If $m_1=0.06\mathrm{kg}$, then the value of $m_2=\left(4.00-0.06\right)\mathrm{kg}=3.94\mathrm{kg}$.

Thus, the masses of the two objects are 0.06 kg and 3.94 kg.