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.
7.2 Kepler's laws
Kepler's laws describe the motion of planets around the Sun based on precise astronomical observations. These laws are:
1. Law of Orbits: All planets move in elliptical orbits with the Sun at one focus of the ellipse. An ellipse is a closed curve defined such that the sum of the distances from any point on the ellipse to the two foci is constant. The closest point of the orbit to the Sun is called the perihelion, and the farthest point is the aphelion. The semi-major axis is half the longest diameter of the ellipse.
2. Law of Areas: The line joining a planet and the Sun sweeps out equal areas in equal intervals of time. This means that a planet moves faster when it is near perihelion and slower near aphelion, conserving angular momentum.
3. Law of Periods: The square of the orbital period (T²) of a planet is proportional to the cube of the semi-major axis (a³) of its orbit. Mathematically, T² ∝ a³. This law allows us to relate the time a planet takes to orbit the Sun to the size of its orbit.
The law of areas can be derived from the conservation of angular momentum, which holds for any central force like gravity. The angular momentum L = r × p (where r is position vector and p is momentum) remains constant, implying that the areal velocity (area swept per unit time) is constant.
Example 7.1 illustrates the relationship between the speed and distance of a planet at perihelion and aphelion using conservation of angular momentum. It shows that the planet moves faster at perihelion than at aphelion and takes longer to traverse the larger area between aphelion and perihelion.
📊 Diagram: Fig. 7.1(a): Elliptical orbit of a planet around the Sun showing perihelion (P), aphelion (A), and semi-major axis. Fig. 7.1(b): Method to draw an ellipse using two foci and a string. Fig. 7.2: Illustration of the area swept by the radius vector in a small time interval Δt.
🧪 Activity: No specific activity but includes worked example applying angular momentum conservation.
🔗 Connection: Prepares for Section 7.3 where Newton's universal law of gravitation is introduced, explaining the force behind Kepler's laws.
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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