Newton's Law of Gravitation, Field and Potential

Every mass attracts every other mass with F = Gm₁m₂/r². From that one law flow the gravitational field, the negative potential, and every satellite equation — the cleanest topic in the JEE syllabus.

Law of Gravitation1 min · Free lecture

In this lesson

State Newton's law of gravitation and the value of G
Compute gravitational field intensity g = GM/r²
Use gravitational potential V = −GM/r and potential energy U = −GMm/r

Two point masses $m_1$ and $m_2$ separated by $r$ attract each other with a force along the line joining them: $F = Gm_1m_2/r^2$ . The constant $G = 6.67 \times 10^{-11}\,\text{N·m}^2/\text{kg}^2$ is universal — the same for apples and planets.

F = \frac{G m_1 m_2}{r^2}, \qquad g = \frac{GM}{r^2}, \qquad V = -\frac{GM}{r}

Gravitational force, field intensity and potential of a point mass M

G — universal gravitational constant (N·m^2/kg^2)
M — mass of the source (kg)
r — distance from the centre of M (m)
g — gravitational field intensity (acceleration due to gravity) (m/s^2)
V — gravitational potential (J/kg)

The field $g$ is the force per unit mass; the potential $V$ is the potential energy per unit mass. Both are defined relative to infinity, where they vanish. Because gravity is attractive, the potential is negative and it increases (becomes less negative) as you move away — a frequent trap in sign-based questions.

U = -\frac{GMm}{r}

Gravitational potential energy of mass m at distance r from M

U — potential energy (negative, zero at infinity) (J)
m — mass in the field of M (kg)

For a spherical body (like Earth), r is measured from the CENTRE, not the surface. g at Earth's surface: g = GM/R² ≈ 9.8 m/s² with R ≈ 6.37 × 10⁶ m.

Worked example

Find the gravitational field intensity at a height h = R above the Earth's surface (R = 6.4 × 10⁶ m, M = 6.0 × 10²⁴ kg, G = 6.67 × 10⁻¹¹).

r = R + h = 2R = 1.28 × 10⁷ m
g = GM/r² = (6.67e-11 × 6.0e24) / (1.28e7)² = 4.0e14 / 1.638e14 ≈ 2.44 m/s²
Check: g' = g/4 = 9.8/4 ≈ 2.45 m/s² ✓ (height R halves gravity to a quarter)

Answer: g ≈ 2.45 m/s²

Transcript (1 min)

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Newton's Law of Gravitation, Field and Potential
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F = G m₁ m₂ / r²
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g = GM / r²
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V = −GM / r — negative, zero at infinity
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r measured from the CENTRE of the sphere
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Example: h = R → g' = g/4 ≈ 2.45 m/s²
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Recap: F = Gm₁m₂/r² · g = GM/r² · V = −GM/r

Printable notes (free)

Practice

Free · 4 questions with full solutions

Q1 · Numerical · difficulty 2/5

Two point masses 44 kg and 1 kg are separated by 2.5 m. Find the gravitational force between them (G = 6.67 × 10⁻¹¹ N·m²/kg²).

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

Q2 · Numerical · difficulty 2/5

Find the gravitational field intensity at distance 1 × 10⁷ m from the centre of a body of mass 7 × 10²⁴ kg (G = 6.67 × 10⁻¹¹ N·m²/kg²).

g = \frac{GM}{r^2}

Q3 · Numerical · difficulty 2/5

Find the escape velocity from the surface of a planet of mass 7 × 10²⁴ kg and radius 4.5 × 10⁶ m (G = 6.67 × 10⁻¹¹ N·m²/kg²).

v_e = \sqrt{\frac{2GM}{R}}

Q4 · MCQ · difficulty 2/5

The gravitational potential due to a point mass at a finite distance from it is:

Always positive
Always negative
Zero
Positive near the mass and negative far away