Gravitation | Class 11 Physics Notes
By ConceptScroll Team · Published on 17 July 2026 · 5 min read
Gravitation – this guide gives you a concise, exam-ready overview of Gravitation from Class 11 Physics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.
7.3 Universal law of gravitation
Newton's universal law of gravitation unifies terrestrial gravity and celestial mechanics. Inspired by the observation of an apple falling and the Moon's orbital motion, Newton proposed that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.
Newton reasoned that the Moon's centripetal acceleration (a_m) due to Earth's gravity is given by a_m = V² / R_m = 4π² R_m / T², where V is the Moon's orbital speed, R_m its orbital radius, and T its orbital period (~27.3 days). This acceleration is much smaller than the acceleration due to gravity (g) on Earth's surface, indicating that gravitational force decreases with distance.
Assuming inverse square dependence, g / a_m = (R_m / R_E)² ≈ 3600, consistent with measured values of g and a_m. This led Newton to state the Universal Law of Gravitation:
"Every body in the universe attracts every other body with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them."
Mathematically, the magnitude of the gravitational force F between two point masses m_1 and m_2 separated by distance r is:
F = G (m_1 m_2) / r²
where G is the universal gravitational constant. The force vector points along the line joining the masses and is attractive.
For extended bodies, the total gravitational force is the vector sum of forces due to all constituent point masses. Two important results simplify calculations:
1. A hollow spherical shell attracts an external point mass as if all its mass were concentrated at its center.
2. The gravitational force inside a hollow spherical shell is zero.
Example 7.2 applies the principle of superposition to calculate the net gravitational force on a mass placed at the centroid of an equilateral triangle formed by three equal masses, showing the resultant force is zero due to symmetry.
📊 Diagram: Fig. 7.3: Vector diagram showing gravitational force on mass m_1 due to m_2 along vector r = r_2 - r_1. Fig. 7.4: Multiple masses exerting gravitational forces on a point mass, illustrating vector addition. Fig. 7.5: Three equal masses at vertices of an equilateral triangle with a mass 2m at the centroid.
🧪 Activity: No specific activity, but includes detailed vector force addition example.
🔗 Connection: Leads to Section 7.4 where the gravitational constant G is experimentally determined.
Frequently asked questions
(a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means? (b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity? (c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun's pull is greater than the moon's pull. (you can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon's pull is greater than the tidal effect of sun. Why?
(a) No, you cannot shield a body from gravitational influence by placing it inside a hollow sphere or by any other means. This is because gravity is always attractive and acts through mass, and there is no known material or configuration that can block gravitational fields. Unlike electrical forces which can be shielded by conductors due to the presence of positive and negative charges, gravity has only one type of 'charge' (mass), so it cannot be shielded.
(b) An astronaut inside a small space
(a) Acceleration due to gravity increases/decreases with increasing altitude. (b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density). (c) Acceleration due to gravity is independent of mass of the earth/mass of the body. (d) The formula $-G M m (1 / r_{2} - 1 / r_{1})$ is more/less accurate than the formula $m g (r_{2} - r_{1})$ for the difference of potential energy between two points $r_{2}$ and $r_{1}$ distance away from the centre of the earth.
(a) Acceleration due to gravity decreases with increasing altitude because gravitational force decreases with distance from the Earth's center.
(b) Acceleration due to gravity decreases with increasing depth inside the Earth (assuming uniform density) because the effective mass attracting the object decreases linearly with depth.
(c) Acceleration due to gravity is independent of the mass of the body; it depends on the mass of the Earth and the distance from its center.
(d) The formula $-G M m
Suppose there existed a planet that went around the Sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?
Let the orbital period of Earth be T and orbital radius be R.
Given: New planet's period T' = T/2.
From Kepler's third law: T^2 ∝ R^3
Therefore, (T')^2 / T^2 = (R')^3 / R^3
=> (T/2)^2 / T^2 = (R')^3 / R^3
=> (1/4) = (R')^3 / R^3
=> (R')^3 = R^3 / 4
=> R' = R / (4)^{1/3} = R / 1.5874 ≈ 0.63 R
So, the orbital radius of the planet is approximately 0.63 times that of the Earth.
Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is $4.22 \times 10^{8} \mathrm{~m}$ . Show that the mass of Jupiter is about one-thousandth that of the sun.
Given: Orbital period, T = 1.769 days = 1.769 × 24 × 3600 = 152,793.6 s Orbital radius, r = 4.22 × 10^8 m
Using the formula for orbital period:
T^2 = \frac{4 \pi^2 r^3}{G M}
Rearranged to find mass M:
M = \frac{4 \pi^2 r^3}{G T^2}
Substitute values:
G = 6.67 × 10^{-11} N·m^2/kg^2
M = \frac{4 \pi^2 (4.22 \times 10^8)^3}{6.67 \times 10^{-11} (152,793.6)^2}
Calculate numerator: (4.22 × 10^8)^3 = 7.52 × 10^{25} m^3
4 π^2 × 7.52 × 10^{25} ≈ 4 × 9.87 × 7.52 × 10^{25} ≈ 296.5 × 10^{25} = 2.965
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