Class 11 Physics · Chapter 7 NotesGravitation

Early models of the universe placed the Earth at the centre (Ptolemy's geocentric theory), with the Sun, Moon and planets moving around it. Aryabhata in the 5th century and later Copernicus proposed a heliocentric picture with the Sun at the centre. Tycho Brahe collected very accurate planetary observations, which Johannes Kepler used to formulate three laws. The first law states that every planet moves in an elliptical orbit with the Sun at one focus. The second law states that the line joining a planet to the Sun sweeps equal areas in equal intervals of time, so a planet moves faster near the Sun (perihelion) and slower far from it (aphelion). The third law states that the square of the orbital period is proportional to the cube of the semi-major axis: T² ∝ a³.

• Perihelionnearest point to the Sun; Aphelion: farthest point from the Sun.
• Semi-major axis ahalf of the longest diameter of the elliptical orbit.
• For a circular orbit, both foci coincide and a equals the radius.
• Kepler's second law follows from conservation of angular momentum.

Newton proposed that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In symbols, F = Gm₁m₂/r², where G is the universal gravitational constant. The force is always attractive, acts along the line joining the two particles, is a central force, and is mutual — both particles experience equal and opposite forces. It obeys an inverse square law and has infinite range. When several masses are present, the net gravitational force on any one mass is the vector sum of the individual forces, which is the principle of superposition.

• F = Gm₁m₂/r²
• G = 6.67 × 10⁻¹¹ N m² kg⁻²
• G was first measured by Henry Cavendish in 1798 using a torsion balance.
• Knowing G allows us to determine the mass of the Earth and of other planets.

For a uniform spherical shell of mass M and radius R, the gravitational force it exerts on a particle of mass m outside the shell is the same as if the entire mass M were concentrated at the centre: F = GMm/r², where r is the distance from the centre. This is true for any point outside the shell. However, for a point inside a hollow spherical shell, the gravitational force is exactly zero. This is an important result: inside a hollow shell, the gravitational field vanishes, but the gravitational potential is not zero. These results are often called the shell theorem and are useful in treating the Earth as a sphere.

• Outside a shellF = GMm/r², as if all mass is at the centre.
• Inside a hollow shellF = 0.
• Inside a hollow shell, gravitational potential is non-zero even though force is zero.

The acceleration produced in a body due to the Earth's gravitational pull is called the acceleration due to gravity, denoted by g. For a body on the surface of the Earth, g = GM/R², where M is the mass of the Earth and R its radius. Its standard value is about 9.8 m s⁻². Importantly, g does not depend on the mass of the falling body. As we go to a height h above the surface, g decreases: g_h = GM/(R+h)². For h much smaller than R, this can be approximated as g_h ≈ g(1 − 2h/R). As we go to a depth d below the surface, g decreases linearly: g_d = g(1 − d/R). At the centre of the Earth, g becomes zero.

• g = GM/R² ≈ 9.8 m s⁻²
• g_h = GM/(R+h)²; for h << R, g_h ≈ g(1 − 2h/R)
• g_d = g(1 − d/R); g = 0 at the Earth's centre.
• g is independent of the mass of the falling body.

The gravitational potential energy of a mass m at a distance r from a point mass M is U = −GMm/r. The negative sign indicates that the force is attractive and that the potential energy is taken as zero at infinity. Gravitational potential V at a point is defined as the potential energy per unit mass: V = −GM/r. Its SI unit is J kg⁻¹. Both U and V are scalar quantities, so when several masses are present, the total potential at a point is the algebraic sum of the potentials due to each mass individually. Gravitational force is conservative, which is why a potential energy function can be defined.

• U = −GMm/r
• V = −GM/r, unit J kg⁻¹
• Potential is zero at infinity and negative elsewhere.
• Potential is a scalar; use algebraic addition for multiple masses.

Escape speed is the minimum speed required for an object to escape from the gravitational influence of a planet without any further propulsion. For the Earth, it is given by v_e = √(2GM/R), which can also be written as v_e = √(2gR). Substituting the values for the Earth gives v_e ≈ 11.2 km s⁻¹. The escape speed depends only on the mass and radius of the planet, not on the mass of the projectile or the direction of launch. This concept explains why the Moon, with an escape speed of about 2.3 km s⁻¹, cannot retain a significant atmosphere: gas molecules can easily exceed this speed.

• v_e = √(2GM/R) = √(2gR)
• For Earth, v_e ≈ 11.2 km s⁻¹
• Independent of the mass and direction of projection.
• Moon's escape speed ≈ 2.3 km s⁻¹, so it has no significant atmosphere.

A satellite revolves around the Earth because the gravitational force provides the necessary centripetal force. For a satellite of mass m in a circular orbit of radius r around the Earth of mass M, we equate mv²/r = GMm/r², giving the orbital speed v = √(GM/r). Near the Earth's surface, this becomes v = √(gR) ≈ 7.9 km s⁻¹. The time period of the satellite is T = 2π√(r³/GM), which shows that T² ∝ r³, consistent with Kepler's third law. For a satellite near the Earth, the period is about 84–85 minutes. The kinetic energy is K = GMm/(2r), the potential energy is U = −GMm/r, and the total energy is E = −GMm/(2r). Note that U = −2K and E = −K. The total energy of a bound satellite is always negative.

• Orbital speedv = √(GM/r); near Earth, v ≈ 7.9 km s⁻¹
• Time periodT = 2π√(r³/GM); near Earth, T ≈ 84–85 minutes
• Kinetic energy K = GMm/(2r)
• Potential energy U = −GMm/r
• Total energy E = −GMm/(2r); E is negative for a bound orbit.

Astronauts in an orbiting spacecraft feel weightless not because gravity is absent, but because both the astronaut and the spacecraft are in a state of continuous free fall towards the Earth. Since they fall together with the same acceleration, the astronaut does not press against the floor of the spacecraft, and the normal reaction becomes zero. This condition is called weightlessness. The gravitational force is still acting, and it is this force that provides the centripetal acceleration needed to keep the spacecraft in orbit. Thus, weightlessness is a consequence of free fall, not of zero gravity.

• Weightlessness occurs because astronaut and spacecraft are in free fall together.
• Gravity is not zero; it provides the centripetal force.
• Normal reaction on the astronaut becomes zero.