Q.The force of gravitation is
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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 m1 and m2 separated by a distance r, the magnitude of the attraction is
F=r2Gm1m2,
where G=6.67×10−11 N m2kg−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=(R2+h2)3/2GMmh, …
Gravitational force can be derived from a potential energy function (U=−rGMm) and the work it does around any closed path is zero, so it is a conserva …
The force of gravitation is a conservative force.
Gravitation is always attractive (never repulsive, ruling out option a) and it is not electrostatic in origin — it acts between masses, not charges (ruling out option b).
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Showing the 12 most recent of 32 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.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
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- CBSE 2026Set ANNUAL1 markMCQQ.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:
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- CBSE 2026Set ANNUAL1 markMCQQ.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
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- CBSE 2026Set ANNUAL1 markMCQQ.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=Gm1m2/r2, doubling r divides F by 22=4.
Newton's law of gravitation gives the force between two point masses as:
F=r2Gm1m2
If the separation is doubled (r→2r), the new force is: …
- CBSE 2026Set ANNUAL1 markQ.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 kg.
Weight is the gravitational force on an object, W=mg, where g is the acceleration due to gravity (≈9.8 m/s2 on Earth).
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- CBSE 2026Set ANNUAL1 markQ.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) …
- CBSE 2026Set ANNUAL1 markMCQQ.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] …
- CBSE 2026Set ANNUAL1 markMCQQ.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:
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- CBSE 2025Set ANNUAL1 markMCQQ.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=r2Gm1m2⇒G=m1m2Fr2
Substituting dimensions: [F]=MLT−2, [r2]=L2, [m1m2]=M2:
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- CBSE 2025Set ANNUAL1 markQ.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 6.67×10−9 N — vanishingly small, which is why gravity between everyday objects is never noticeable.
Newton's law of gravitation:
F=Gm1m2/r2
Here: G=6.674×10−11 N·m²/kg², m1=m2=20 kg, r=2 m
F=(6.674×10−11)×(20×20)/(2)2
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- CBSE 2025Set ANN1 markQ.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.
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- CBSE 2024Set ANNUAL1 markMCQQ.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/R2.
Acceleration due to gravity at the earth's surface: g=R2GM, where M is earth's mass and R its radius. Dividing both sid …
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