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.
7.1 Introduction
From early childhood, we observe that all material objects tend to be attracted towards the Earth. This is evident in everyday phenomena such as objects falling when thrown up, raindrops descending from clouds, and the greater effort required to climb uphill compared to downhill. The Italian physicist Galileo Galilei (1564–1642) was the first to recognize that all bodies, regardless of their masses, accelerate towards the Earth with the same constant acceleration. He demonstrated this fact experimentally, notably by studying bodies rolling down inclined planes, and estimated the acceleration due to gravity, which was later measured more accurately.
Simultaneously, the motion of celestial bodies like stars and planets has fascinated humans since ancient times. Early observations noted that some stars maintained fixed positions year after year, while planets exhibited regular motions against this fixed background. The earliest recorded planetary motion model was the geocentric model proposed by Ptolemy about 2000 years ago, which placed Earth at the center with all celestial bodies revolving around it in circular orbits. To explain observed planetary motions, Ptolemy introduced complex schemes involving circles upon circles.
Indian astronomers also developed similar models about 400 years later. However, Aryabhatta, in the 5th century A.D., proposed a heliocentric idea where the Sun was at the center with planets revolving around it. This idea was definitively formulated by Nicolaus Copernicus (1473–1543), who proposed circular orbits around the Sun. Despite opposition from the church, Galileo supported the heliocentric model, facing prosecution for his beliefs.
Around the same time, Tycho Brahe (1546–1601) made extensive naked-eye observations of planetary positions. His assistant Johannes Kepler (1571–1640) analyzed this data and formulated three fundamental laws of planetary motion, now known as Kepler's laws. These laws were crucial for Newton's formulation of the universal law of gravitation, which unified terrestrial and celestial mechanics.
📊 Diagram: No specific diagrams in this introductory section, but historical models of planetary motion are referenced.
🧪 Activity: No activity in this section.
🔗 Connection: Leads into Section 7.2 where Kepler's laws are discussed in detail, providing the foundation for understanding planetary motion and gravitation.
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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