Discuss the important features of the law of gravitation.

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Q. Discuss the important features of the law of gravitation.

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Concept understanding — Newton's Law of Universal Gravitation

Newton's Law of Universal Gravitation

Every particle of matter attracts every other particle with a force directed along the line joining them. For two point masses $m_1$ and $m_2$ separated by a distance $r$ , the magnitude of the attraction is

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

where $G = 6.67 \times 10^{-11}\ \text{N m}^{2}\,\text{kg}^{-2}$ is the universal gravitational constant. The force is attractive, acts along the line of centres, and obeys Newton's third law (equal and opposite on the two bodies). Its defining feature is the inverse-square dependence on separation.

Superposition. When several masses act on a body, the net gravitational force is the vector sum of the individual forces. For an extended body — a rod, ring or shell — the body is treated as a collection of point-mass elements and the contributions are integrated. For a uniform ring of mass $M$ and radius $R$ , the force on a point mass $m$ placed on its axis at distance $h$ from the centre is directed along the axis and has magnitude

F = \frac{G M m\, h}{\left(R^{2} + h^{2}\right)^{3/2}},

because the components perpendicular to the axis cancel by symmetry while the axial components add. This axial formula is not a simple $1/h^{2}$ law: as $h$ changes, the force scales as $h / (R^{2}+h^{2})^{3/2}$ , so comparing the force at two axial positions (say $h$ and $2h$ ) means substituting into this expression rather than using the point-mass inverse square directly. Only when $h \gg R$ does the ring behave like a point mass and $F \to GMm/h^{2}$ .

Newton's law of gravitation is one of the earliest topics in the NCERT Class 11 Physics chapter on Gravitation, and "universal law of gravitation formula and examples" or "gravitation class 11 important questions" are searched constantly by CBSE board and NEET/JEE Main aspirants. The extended-body superposition case shown here — finding the force due to a ring on an axial point mass — is a classic JEE Main gravitation numerical built directly on this NCERT Class 11 foundation.

Step 1. Inverse-square dependence. $F=GM_1M_2/r^2$ falls off rapidly with distance -- doubling the separation cuts the force to a quarter. This is why a far planet like Uranus feels a much weaker solar pull than Earth does.

Step 2. Always attractive. The force always points along the line joining the two masses, back toward the other mass -- gravity never repels, unlike the electric force.

Step 3. Action-reaction pair. By Newton's third law, the force $M_1$ exerts on $M_2$ is equal in magnitude and opposite in direction to the force $M_2$ exerts on $M_1$ : the Sun pulls Earth toward itself, and Earth pulls the Sun toward itself with an equal force, even though the Sun barely moves because of its enormous mass.

Step 4. Point-mass assumption. The formula strictly applies to point masses, or to any two bodies whose separation is far larger than their own physical sizes -- an excellent approximation for planet-Sun systems.

Step 5. The hollow-sphere exception. One extended-body case works out exactly: a uniform hollow sphere of mass $M$ attracts an external point mass exactly as if all of $M$ were concentrated at the sphere's centre; and a mass placed inside such a shell feels exactly zero net force, from anywhere inside it.

Step 6. Extremely weak in ordinary situations. Because $G\approx6.67\times10^{-11}$ is such a tiny number, gravity between two everyday-sized objects is far too small to notice -- only once one mass is planet-sized does gravity become dominant.

✓Final answer

G

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 32 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Who formulated the law of universal gravitation?$
(a) Albert Einstein
(b) Galileo Galilei
(c) Isaac Newton
(d) Johannes Kepler
›Reveal solutionSolution

Isaac Newton formulated the law of universal gravitation, stating that every particle attracts every other particle with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.

While Johannes Kepler discovered the empirical laws of planetary motion (Kepler's three laws), it was Isaac Newton who generalised these observations into a single universal law of gravitation:

F = G m1 m2 / r^2

where G is the universal gravitational constant. Newton showed that this same law explains both falling objects on Earth and the orbital motion of planets -- hence "universal" gravitation. Galileo studied falling bodies and Einstein later generalised gravity via general relativity, but the law of UNIVERSAL gravitation itself is Newton's.

✓Final answer
$(c) Isaac Newton.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. If the distance between two objects is doubled, how does the gravitational force between them change?$
(a) It becomes half
(b) It becomes one-fourth
(c) It remains the same
(d) It becomes twice
›Reveal solutionSolution

F is proportional to 1/r^2 (inverse-square law); doubling r reduces F to 1/4 of its original value.

Newton's law of gravitation: F = G m1 m2 / r^2

If the distance is doubled, r' = 2r:

F' = G m1 m2 / (2r)^2 = G m1 m2 / (4 r^2) = F/4

So the new force is one-fourth of the original.

✓Final answer
$(b) It becomes one-fourth.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. The gravitational force between two objects is F. If masses of both the objects are halved without altering the distance between them, the gravitational force would become$
(a) F/4
(b) F/2
(c) F
(d) 2F
›Reveal solutionSolution

F is proportional to m1 x m2; halving each mass multiplies the force by (1/2)(1/2) = 1/4.

Newton's law of gravitation: F = G m1 m2 / r^2

If both masses are halved (m1' = m1/2, m2' = m2/2) while r stays the same:

F' = G (m1/2)(m2/2) / r^2 = (1/4) x [G m1 m2 / r^2] = F/4

✓Final answer
$(a) F/4.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. If distance between two bodies is doubled, what is the effect on gravitational force between two bodies:$
(a) become half
(b) become one fourth
(c) become double
(d) become four times
›Reveal solutionSolution

Since $F=Gm_1m_2/r^2$ , doubling $r$ divides $F$ by $2^2=4$ .

Newton's law of gravitation gives the force between two point masses as:

$F = \frac{Gm_1m_2}{r^2}$

If the separation is doubled ( $r\to2r$ ), the new force is:

$F' = \frac{Gm_1m_2}{(2r)^2} = \frac{Gm_1m_2}{4r^2} = \frac{F}{4}$

So the gravitational force becomes one-fourth of its original value.

✓Final answer
$(b) become one fourth.$
CBSE 2026Set ANNUAL1 mark
$Q. What is the mass of an object, whose weight is 98 N on the earth? OR Define Universal Gravitational Constant (G).$
›Reveal solutionSolution

$m=W/g=98/9.8=10\text{ kg}$ .

Weight is the gravitational force on an object, $W=mg$ , where $g$ is the acceleration due to gravity ( $\approx9.8\text{ m/s}^2$ on Earth).

Given $W=98\text{ N}$ :

$m = \frac{W}{g} = \frac{98}{9.8} = 10\text{ kg}$

✓Final answer
$m = 10\text{ kg}$
CBSE 2026Set ANNUAL1 mark
$Q. Write the answer in one word/one sentence: Write the C.G.S. value of universal gravitational constant G.$
›Reveal solutionSolution
G ≈ 6.674 × 10⁻⁸ dyne·cm²/g² in the C.G.S. system.

In SI units: G = 6.674 × 10⁻¹¹ N·m²/kg²

Convert each part to C.G.S.:

1 N = 10⁵ dyne

1 m² = 10⁴ cm²

1 kg² (in the denominator) = (1000 g)² = 10⁶ g²

So: G = 6.674 × 10⁻¹¹ × (10⁵ dyne) × (10⁴ cm²) / (10⁶ g²)

= 6.674 × 10⁻¹¹ × 10⁹⁻⁶ dyne·cm²/g²

= 6.674 × 10⁻⁸ dyne·cm²/g²

✓Final answer
$G \approx 6.674 \times 10⁻⁸ dyne\cdotcm²/g² (commonly rounded to 6.67 \times 10⁻⁸ C.G.S. units).$
CBSE 2026Set ANNUAL1 markMCQ
$Q. The dimensional formula of the universal gravitational constant is$
(a) [M^-1 L^3 T^2]
(b) [M^-1 L^3 T^-1]
(c) [M^-1 L^3 T^-2]
(d) [ML^2 T^-3]
›Reveal solutionSolution
G has dimensions [M^-1 L^3 T^-2]. Answer (C).

Newton's law of gravitation: F = G m1 m2 / r^2, so G = F r^2 /(m1 m2).

Dimensions:

F -> [MLT^-2]

r^2 -> [L^2]

m1 m2 -> [M^2]

G = [MLT^-2][L^2]/[M^2] = [M^(1-2) L^(1+2) T^-2] = [M^-1 L^3 T^-2].

✓Final answer
$(C) [M^-1 L^3 T^-2].$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Two artificial satellites A and B are revolving around a planet in circular orbits of radii 9R and R respectively. If speed of A is 2v, then the speed of B will be$
(a) 2v
(b) 4v
(c) 6v
(d) 9v
›Reveal solutionSolution

v proportional to 1/sqrt(r) gives v_B = 6v. Answer (C).

For a circular orbit, orbital speed v = sqrt(GM/r), so v is proportional to 1/sqrt(r).

Taking the ratio for the two satellites:

v_B/v_A = sqrt(r_A/r_B) = sqrt(9R/R) = sqrt(9) = 3.

Given v_A = 2v, we get v_B = 3 x 2v = 6v.

✓Final answer
$(C) 6v.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Dimension of gravitational constant is (A) M^-1L^3T^-2 (B) ML^3T^-2 (C) M^3L^3T^3 (D) LT^-2$
›Reveal solutionSolution

The gravitational constant G has dimensional formula M⁻¹L³T⁻².

From Newton's law of gravitation:

$F = \dfrac{Gm_1m_2}{r^2} \Rightarrow G = \dfrac{Fr^2}{m_1m_2}$

Substituting dimensions: $[F] = MLT^{-2}$ , $[r^2] = L^2$ , $[m_1m_2] = M^2$ :

$[G] = \dfrac{[MLT^{-2}][L^2]}{[M^2]} = \dfrac{ML^3T^{-2}}{M^2} = M^{-1}L^3T^{-2}$

✓Final answer
$(A) M⁻¹L³T⁻².$
CBSE 2025Set ANNUAL1 mark
$Q. Two 20 kg masses are separated by a distance of 2 metre. Gravitational force between them is .............. .$
›Reveal solutionSolution

Plugging the two 20 kg masses and 2 m separation into Newton's law of gravitation gives a force of about $6.67 \times 10^{-9}$ N — vanishingly small, which is why gravity between everyday objects is never noticeable.

Newton's law of gravitation:

$F = Gm_1m_2/r^2$

Here: $G = 6.674 \times 10^{-11}$ N·m²/kg², $m_1 = m_2 = 20$ kg, $r = 2$ m

$F = \left(6.674 \times 10^{-11}\right) \times \left(20 \times 20\right) / (2)^2$

$F = \left(6.674 \times 10^{-11}\right) \times 400 / 4 = \left(6.674 \times 10^{-11}\right) \times 100$

$F = 6.674 \times 10^{-9}$ N

✓Final answer
$\approx 6.674 \times 10^{-9}$
CBSE 2025Set ANN1 mark
$Q. Acceleration due to gravity is independent of ______ . (mass of earth / mass of body)$
›Reveal solutionSolution

g depends on the Earth's mass and radius, not on the mass of the falling body, so it is independent of the mass of the body.

For a body of mass m at the Earth's surface, the gravitational force is F = GMm/R squared, where M is the Earth's mass and R its radius. The acceleration this force produces is

a = F/m = GMm/(R squared . m) = GM/R squared = g.

The body's own mass m cancels out. Hence g is fixed by M and R (the source of the field), and every object at that location falls with the same acceleration regardless of its mass.

✓Final answer
$Acceleration due to gravity is independent of the mass of the body.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. The radius and mass of the earth are R and M respectively. The ratio of acceleration due to gravity g and G will be (A) R/M (B) R^2/M (C) M/R^2 (D) R^2M$
›Reveal solutionSolution

$g/G=M/R^2$ .

Acceleration due to gravity at the earth's surface: $g=\dfrac{GM}{R^2}$ , where $M$ is earth's mass and $R$ its radius. Dividing both sides by $G$ : $\dfrac{g}{G}=\dfrac{M}{R^2}$ .

✓Final answer
$M/R^2$

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Q.Discuss the important features of the law of gravitation.

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