Why is gravitational potential always negative?

Gravitational potential is defined as the work done by an external agent to bring a unit mass from infinity to a point without acceleration. Since gravity is an attractive force, it naturally pulls masses together. To prevent acceleration, the external agent must apply a force opposite to the gravitational pull, effectively doing negative work. This negative sign signifies that the unit mass is 'bound' to the source mass; energy must be supplied to the system to overcome this binding and move the mass away to infinity. A more negative potential indicates a stronger binding.

What is the difference between gravitational potential and gravitational potential energy?

Gravitational potential ($V$) is a property of the gravitational field at a point, defined as the work done per unit mass (J/kg) to bring a unit mass from infinity to that point. It's independent of the mass placed there. Gravitational potential energy ($U$), on the other hand, is the energy possessed by a specific mass ($m$) due to its position in a gravitational field. It is given by $U = mV$ and is measured in Joules (J). So, potential energy is specific to a mass, while potential describes the field itself.

Is gravitational potential a scalar or vector quantity?

Gravitational potential is a scalar quantity. This means it only has magnitude and no direction. This is a significant advantage when dealing with systems of multiple masses, as the total potential at a point due to several masses can be found by simply algebraically adding the potentials due to each individual mass. This is much simpler than vector addition required for gravitational field intensity.

What is the gravitational potential at the center of a solid sphere?

For a solid sphere of mass $M$ and radius $R$, the gravitational potential at its center ($r=0$) is $V_{\text{center}} = -\frac{3GM}{2R}$. This is the most negative (minimum) value of potential within or outside the sphere. It's more negative than the potential at the surface ($V_{\text{surface}} = -\frac{GM}{R}$), indicating that more work is done by the field to bring a unit mass to the center compared to the surface.

How does gravitational potential relate to escape velocity?

Escape velocity ($v_{\text{esc}}$) is the minimum speed an object needs to be projected from a point so that it can completely escape the gravitational field of a massive body and reach infinity with zero kinetic energy. This concept is directly linked to gravitational potential. If an object of mass $m$ is at a point with potential $V$, its potential energy is $mV$. For it to escape, its initial kinetic energy must be at least equal to the magnitude of its potential energy. Thus, $\frac{1}{2}mv_{\text{esc}}^2 + mV = 0$, which simplifies to $v_{\text{esc}} = \sqrt{-2V}$. The negative sign of $V$ ensures a real value for $v_{\text{esc}}$.

Gravitational Potential

Physics

NEET UG

Version 1Updated 22 Mar 2026

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Gravitational potential at a point in a gravitational field is defined as the amount of work done by an external agent in bringing a unit test mass from infinity to that point without any acceleration. It is a scalar quantity and is denoted by $V$ . Since the gravitational force is attractive, work is done by the field, or equivalently, negative work is done by an external agent. Therefore, gravita…

Quick Summary

Gravitational potential ( $V$ ) at a point is a scalar quantity representing the work done by an external agent to bring a unit test mass from infinity to that point without acceleration. Its SI unit is J/kg.

By convention, potential at infinity is zero, and due to the attractive nature of gravity, gravitational potential is always negative, indicating a bound system. The more negative the potential, the stronger the binding.

For a point mass $M$ at distance $r$ , $V = -GM/r$ . For a spherical shell of radius $R$ and mass $M$ , $V = -GM/r$ for $r \ge R$ and $V = -GM/R$ for $r < R$ . For a solid sphere of radius $R$ and mass $M$ , $V = -GM/r$ for $r \ge R$ and $V = -\frac{GM}{2R^3}(3R^2 - r^2)$ for $r < R$ .

The potential at the center of a solid sphere is $-3GM/(2R)$ . Gravitational potential energy ( $U$ ) of a mass $m$ at a point is $U = mV$ . The gravitational field intensity $\vec{E}$ is related to potential by $\vec{E} = -\nabla V$ .

This concept is crucial for understanding energy in gravitational fields and phenomena like escape velocity.

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Key Concepts

Gravitational Potential due to a Point Mass

The most fundamental case is the potential created by a single, isolated point mass $M$ . At any distance $r$ …

Gravitational Potential due to a Spherical Shell

For a thin spherical shell of mass $M$ and radius $R$ , the gravitational potential behaves differently inside…

Gravitational Potential due to a Solid Sphere

For a solid sphere of uniform density, mass $M$ , and radius $R$ , the potential outside ( $r \ge R$ ) is again…

Definition: — Work done by external agent to bring unit mass from

\infty

to point (J/kg).\n- Sign: Always negative (attractive force, bound system).\n- Scalar: Yes, add algebraically.\n- Point Mass:

V = -\frac{GM}{r}

\n- **Spherical Shell (Mass

M

, Radius

R

):**\n - Outside (

r \ge R

V = -\frac{GM}{r}

\n - Inside (

r < R

V = -\frac{GM}{R}

(constant)\n- **Solid Sphere (Mass

M

, Radius

R

):**\n - Outside (

r \ge R

V = -\frac{GM}{r}

\n - Inside (

r < R

V = -\frac{GM}{2R^3}(3R^2 - r^2)

\n - Center (

r=0

V_{\text{center}} = -\frac{3GM}{2R}

\n- Relation to Potential Energy:

U = mV

\n- Relation to Field Intensity:

\vec{E} = -\nabla V

(Field points towards decreasing potential)\n- Escape Velocity:

v_{\text{esc}} = \sqrt{-2V}

Very Negative Gravity Makes Radius Small. (V = -GM/r) - Helps recall the point mass formula and negative sign. \n\nSolid Sphere Center 3/2 Surface. (V_center = 3/2 * V_surface) - Reminds the relation between center and surface potential for a solid sphere.