Question

Two identical objects each of mass $50 \,kg$ are kept at a distance of separation of $50\, cm$ apart on a horizontal table. The net gravitational force at the mid-point of the line joining their centres is

• A. zero
• B. $6.6733 \times 10^{-9}N$
• C. $ 13.346 \times 10^{-9}N$
• D. $ 3.336 \times 10^{-9}N$

Answer 1

The Correct Option is A

The net gravitational force between two objects depends on their masses and the distance between them. The formula for gravitational force is given by :
\(F = \frac{G \times (m_1 \times m_2) }{ r^2}\)
where G is the gravitational constant, m₁ and m₂ are the masses of the two objects, and r is the distance between them.
In this case, the two objects are identical and have the same mass, m₁ = m₂ = 50 kg.
The distance between them is r = 50 cm = 0.5 m.
The gravitational force between the two objects is:

\(F = \frac{G \times (m1 \times m2) }{r^2} = \frac{6.6733×10^{−11} \times (50 \times 50)} {(0.5 \times 0.5)} = 6.6733×10^{−9} N\)
The gravitational force on each object due to the other is equal in magnitude and opposite in direction. Therefore, the net gravitational force at the mid-point of the line joining their centres is zero.
Hence, the correct answer is zero.
So, the correct option is (A) : Zero.

Answer 2

Given :
m = 50 kg, 2r = 50 cm.
If the mass of each object is m and and the distance between two objects is 2r
So, The gravitational force at the mid-point of the line joining the centre due to both the object will be equal and the opposite.
∴ \(\vec{F}=\vec{F_1}+\vec{F_2}\)
\(=G\frac{m}{r^2}\hat{r}+G\frac{m}{r^2}-\hat{r}=0\)
So, the correct option is (A) : Zero.