Physics 2021 JAMB Past Questions - Question 31: Two equal masses placed 2m apart, have a gravitational attraction of 6.7 × 10¹N. compute the value of the mass. [Assume, G = 6.7 × 10-¹¹Nm²/kg²].

Correct Answer

A

Explanation

The gravitational force between two masses can be calculated using Newton's law of gravitation, which is given by the formula:

\[ F = \frac{{G \cdot m_1 \cdot m_2}}{{r^2}} \]

where:
- \( F \) is the gravitational force,
- \( G \) is the gravitational constant (\(6.7 \times 10^{-11} \, \text{N m}^2/\text{kg}^2\)),
- \( m_1 \) and \( m_2 \) are the masses,
- \( r \) is the separation between the centers of the masses.

In this case, you're given that the gravitational force (\( F \)) is \(6.7 \times 10^9 \, \text{N}\), the separation (\( r \)) is \(2 \, \text{m}\), and you want to find the mass (\( m \)).

Substitute the known values into the formula:

\[ 6.7 \times 10^9 = \frac{{(6.7 \times 10^{-11}) \cdot m \cdot m}}{{(2)^2}} \]

Now, solve for \( m \):

\[ 6.7 \times 10^9 = \frac{{6.7 \times 10^{-11} \cdot m^2}}{{4}} \]

Multiply both sides by 4:

\[ 4 \cdot 6.7 \times 10^9 = 6.7 \times 10^{-11} \cdot m^2 \]

\[ 2.68 \times 10^{10} = 6.7 \times 10^{-11} \cdot m^2 \]

Now, solve for \( m \):

\[ m^2 = \frac{{2.68 \times 10^{10}}}{{6.7 \times 10^{-11}}} \]

\[ m^2 = 4 \times 10^{20} \]

\[ m = \sqrt{4 \times 10^{20}} \]

\[ m = 2 \times 10^{10} \, \text{kg} \]

So, the mass of each object is \(2 \times 10^{10} \, \text{kg}\).