Physics Q16 – NEET-UG 2023

Two bodies of mass

m

and

9 m

are placed at a distance

R

. The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be (

G = g r a v i t a t i o n a l c o n s t a n t

Clue 1 of 3: Gravitational Field and Potential

Understand that the gravitational field is a vector quantity, and potential is a scalar quantity.

🧠 Full Solution Path

Step 1: Locate the point of zero gravitational field✦ Active

Let the point where the gravitational field is zero be at a distance $x$ from mass $m$ and $R - x$ from mass $9 m$ . For the net field to be zero, the magnitudes of the fields due to each mass must be equal:

\frac{G m}{x^{2}} = \frac{G (9 m)}{(R - x)^{2}}

Taking the square root on both sides (and considering the point is between the masses):

\frac{1}{x} = \frac{3}{R - x} ⟹ R - x = 3 x ⟹ R = 4 x ⟹ x = \frac{R}{4}

So, the point is at a distance $\frac{R}{4}$ from mass $m$ and $R - \frac{R}{4} = \frac{3 R}{4}$ from mass $9 m$ .

Step 2: Calculate the gravitational potential at this point○ Expand

The total gravitational potential $V$ at this point is the scalar sum of the potentials due to each mass:

V = V_{m} + V_{9 m} = - \frac{G m}{x} - \frac{G (9 m)}{R - x}

💡 Teacher's Secret Hint

Remember that gravitational potential is always negative and is a scalar quantity, so we simply add the potentials.

Step 3: Substitute the value of

x

and simplify○ Expand

Substitute $x = \frac{R}{4}$ and $R - x = \frac{3 R}{4}$ into the potential equation:

V = - \frac{G m}{R / 4} - \frac{G (9 m)}{3 R / 4}

Simplify the expression:

V = - \frac{4 G m}{R} - \frac{4 \times 9 G m}{3 R} = - \frac{4 G m}{R} - \frac{36 G m}{3 R} = - \frac{4 G m}{R} - \frac{12 G m}{R}

Combine the terms:

V = - \frac{(4 + 12) G m}{R} = - \frac{16 G m}{R}

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Physics Question 16 – NEET-UG 2023

Clue 1 of 3: Gravitational Field and Potential

🧠 Full Solution Path