Sigma Percentile
LEVELJEE Main

Animated Solution for Physics - Gravitation: Two bodies of masses and are placed at a distance . The gravitational potential at a point on the line joining them, where the gravitational field is zero, is

Select Answer:

Visualized Solution

  • Let the neutral point be at a distance from mass .

  • At neutral point , the gravitational fields must cancel out.

  • Gravitational potential is a scalar quantity.

  • Substitute and

  • What if the masses were solid spheres instead of point masses?

The Sigma Insight: Gravitational Potential and Potential Energy

Solution Diagram

The Setup

Finding the Neutral Point
Imagine two masses, and , separated by a distance . We are tasked with finding a specific point on the line joining them where the net gravitational field is exactly zero. This point is often called the neutral point.
Let's assume this point is located at a distance from the smaller mass . Consequently, its distance from the larger mass will be .

Balancing the Fields

For the net gravitational field to be zero at point , the gravitational pull from mass must perfectly cancel out the pull from mass . Since these two forces act in opposite directions along the line joining the masses, their magnitudes must be equal.
We can set up the equation by equating the magnitudes of the gravitational fields:

Solving for the Position

Now, let's solve for . We can start by canceling the common terms, and , from both sides of the equation:
Taking the square root of both sides simplifies things beautifully. Since lies between the masses, we only consider the positive root:
Cross-multiplying gives us a simple linear equation:
So, the neutral point is located at a distance of from mass , and from mass .

Calculating the Gravitational Potential

Now that we have the exact location of point , we need to find the total gravitational potential there. Remember, unlike the gravitational field, gravitational potential is a scalar quantity. This means we don't need to worry about directions; we just add the potentials due to each mass algebraically.
The total potential at point is the sum of the potential due to and the potential due to :

The Final Result

Let's substitute our value of into the potential equation:
Simplifying the fractions, we get:
Adding these two terms together yields our final answer:
This is the total gravitational potential at the point where the gravitational field is zero. It's a beautiful demonstration of how vector fields and scalar potentials interact in a simple two-body system!

Similar Questions

JEE Main 2020, 4 Sep Shift-I
LEVELJEE Main

On the X-axis and at a distance from the origin, the gravitational field due to a mass distribution is given by in the -direction. The magnitude of gravitational potential on the X-axis at a distance , taking its value to be zero at infinity, is

(A)
(B)
(C)
(D)
JEE Main 2021, 27 Aug Shift-II
LEVELJEE Main

A mass of is placed at the centre of a uniform spherical shell of mass and radius . If the gravitational potential at a point, from the centre is . The value of is

(A)
(B)
(C)
(D)
JEE Main 2015
LEVELJEE Advanced

From a solid sphere of mass and radius , a spherical portion of radius is removed as shown in the figure. Taking gravitational potential at , the potential at the centre of the cavity thus formed is ( gravitational constant)

(A)
(B)
(C)
(D)
JEE Advanced 1993
LEVELJEE Advanced

A solid sphere of uniform density and radius units is located with its centre at the origin of coordinates. Two spheres of equal radii unit, with their centres at and respectively, are taken out of the solid leaving behind spherical cavities as shown in figure.

* Multiple Correct Options
(A)
the gravitational field due to this object at the origin is zero
(B)
the gravitational field at the point is zero
(C)
the gravitational potential is the same at all points of circle
(D)
the gravitational potential is the same at all points on the circle
JEE Advanced 2013
LEVELJEE Advanced

Two bodies, each of mass , are kept fixed with a separation . A particle of mass is projected from the mid-point of the line joining their centres, perpendicular to the line. The gravitational constant is . The correct statement(s) is (are)

* Multiple Correct Options
(A)
The minimum initial velocity of the mass to escape the gravitational field of the two bodies is
(B)
The minimum initial velocity of the mass to escape the gravitational field of the two bodies is
(C)
The minimum initial velocity of the mass to escape the gravitational field of the two bodies is
(D)
The energy of the mass remains constant
JEE Main 2021, 27 Aug Shift-I
LEVELJEE Advanced

A body of mass splits into four masses , which are rearranged to form a square as shown in the figure. The ratio of for which, the gravitational potential energy of the system becomes maximum is . The value of is……… .

LEVELJEE Main

If is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass raised from the surface of the earth to a height equal to the radius of the earth, is

(A)
(B)
(C)
(D)
JEE Advanced 1983
LEVELJEE Main

If is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass raised from the surface of the earth to a height equal to the radius of the Earth, is

(A)
(B)
(C)
(D)
JEE Main 2021, 27 July Shift-I
LEVELJEE Advanced

Suppose two planets (spherical in shape) of radii and , but mass and respectively have a centre to centre separation as shown in the figure. A satellite of mass is projected from the surface of the planet of mass directly towards the centre of the second planet. The minimum speed required for the satellite to reach the surface of the second planet is , then the value of is …………… . [Take, the two planets are fixed in their position]

LEVELJEE Main

Energy required to move a body of mass from an orbit of radius to is

(A)
(B)
(C)
(D)