Gravitation — Class 11 Physics

1. Universal Law of Gravitation

Every two masses attract each other with a force along the line joining their centres. Its magnitude is F = Gm₁m₂/r², where G is the universal gravitational constant. The force is always attractive.

The inverse-square dependence means that if the separation is doubled, the force becomes one-fourth; if separation is tripled, it becomes one-ninth.

2. Gravitational Field

The gravitational field at a point is the gravitational force experienced per unit mass placed at that point: g = F/m. For a spherical mass M, outside the sphere, g = GM/r², directed toward the centre.

3. Acceleration Due to Gravity

At Earth’s surface, g = GM_E/R_E². The acceleration due to gravity is independent of the mass of a freely falling test object when air resistance is neglected.

4. Variation of g with Height

At height h above Earth’s surface, g_h=GM/(R+h)². Therefore g_h=g[R/(R+h)]². For h much smaller than R, the approximation g_h≈g(1−2h/R) is useful.

5. Variation of g with Depth

For an ideal Earth model of uniform density, at depth d below the surface, g_d≈g(1−d/R). Thus g decreases approximately linearly with depth and becomes zero at the centre in this simplified model.

6. Gravitational Potential Energy

Taking zero potential energy at infinity, the gravitational potential energy of mass m at distance r from a mass M is U=−GMm/r. The negative sign indicates that a bound gravitational system has lower energy than the state of infinite separation.

7. Gravitational Potential

Gravitational potential is potential energy per unit mass: V=U/m=−GM/r. It is a scalar quantity.

8. Escape Speed

Escape speed is the minimum speed required for an object to reach infinity with zero final speed, ignoring atmosphere and other bodies. From energy conservation, v_e=√(2GM/R)=√(2gR). It does not depend on the mass of the escaping object.

9. Orbital Velocity

For a circular orbit of radius r, gravitational force provides centripetal force: GMm/r²=mv²/r. Therefore v_o=√(GM/r). Near Earth’s surface, ignoring Earth’s rotation, v_o=√(gR).

10. Time Period of a Satellite

For a circular satellite orbit, T=2π√(r³/GM). Hence T² is proportional to r³, which is the basis of Kepler’s third law for circular orbits.

11. Geostationary Satellite

A geostationary satellite has an orbital period equal to Earth’s rotation period, moves in the equatorial plane in the same direction as Earth’s rotation, and appears fixed over one point on Earth’s surface. Its orbital radius is determined by the condition T≈24 h using the ideal circular-orbit model.

12. Kepler’s Laws

Planets move in elliptical orbits with the Sun at one focus.
The line joining a planet and the Sun sweeps equal areas in equal times.
For planets orbiting the same central body, T² is proportional to a³, where a is the semi-major axis.

Worked Numerical 1 — Gravitational Force

Question: Two masses of 5 kg and 10 kg are separated by 2 m. Take G=6.67×10⁻¹¹ N m²/kg². Find the gravitational force.

F=Gm₁m₂/r²=(6.67×10⁻¹¹)(5)(10)/4=8.34×10⁻¹⁰ N.

Worked Numerical 2 — Variation of g with Height

Question: At a height equal to Earth’s radius R above the surface, what is the value of g in terms of surface g?

g_h=g[R/(R+h)]². With h=R, g_h=g(R/2R)²=g/4.

Worked Numerical 3 — Escape Speed

Question: If g=9.8 m/s² and Earth’s radius is 6.4×10⁶ m, estimate the escape speed.

v_e=√(2gR)=√[2(9.8)(6.4×10⁶)]≈11.2 km/s.

Worked Numerical 4 — Orbital Speed

Question: Near Earth’s surface, estimate the circular orbital speed using g=9.8 m/s² and R=6.4×10⁶ m.

v_o=√(gR)=√[(9.8)(6.4×10⁶)]≈7.9 km/s.

Worked Numerical 5 — Gravitational Potential Energy

Question: Find the gravitational potential energy of a 2 kg mass at a distance 4 m from a 10 kg mass. Take G=6.67×10⁻¹¹ SI units and zero potential energy at infinity.

U=−GMm/r=−(6.67×10⁻¹¹)(10)(2)/4=−3.34×10⁻¹⁰ J.

Worked Numerical 6 — Satellite Period Relationship

Question: Two satellites orbit the same planet in circular orbits of radii r and 4r. Find the ratio of their periods.

Since T∝r^3/2, T₂/T₁=(4r/r)^3/2=8. Therefore T₂:T₁=8:1.

Important Derivation — Escape Speed

For the minimum escape condition, initial kinetic energy at the surface equals the increase in gravitational potential energy required to reach infinity: ½mv_e²=GMm/R. Cancelling m gives v_e=√(2GM/R). This shows why escape speed is independent of the object’s mass.

Important Derivation — Orbital Velocity

For a circular orbit, gravitational attraction supplies centripetal force: GMm/r²=mv²/r. Cancelling m and solving gives v=√(GM/r). Therefore orbital speed decreases as orbital radius increases.

Explained MCQs

If the distance between two masses is doubled, gravitational force becomes:
Answer: One-fourth. F∝1/r².
What happens to g at a height equal to Earth’s radius?
Answer: It becomes g/4. The distance from Earth’s centre becomes 2R.
Escape velocity from Earth depends on:
Answer: Earth’s mass and radius, not the escaping object’s mass. v_e=√(2GM/R).
Which is greater near Earth’s surface: escape speed or circular orbital speed?
Answer: Escape speed. v_e=√2 v_o.
Why is gravitational potential energy negative when zero is chosen at infinity?
Answer: A bound mass has lower energy than at infinite separation. Positive work must be supplied to take it from a finite distance to infinity.
What remains constant for a satellite in a circular orbit?
Answer: Its speed and orbital radius. Its velocity vector is continuously changing direction, so velocity itself is not constant.

Competency Question

Two satellites orbit Earth at different heights. The higher satellite moves more slowly in its circular orbit but takes longer to complete one revolution. Explain both observations using v_o=√(GM/r) and T=2π√(r³/GM).

HOTS

An astronaut inside an orbiting spacecraft appears weightless even though Earth’s gravitational field at the spacecraft’s altitude is not zero. Explain why “weightlessness” in orbit should not be interpreted as “absence of gravity.”

Common Mistakes

Confusing gravitational force with gravitational field.
Using surface g at a large altitude without checking the distance from Earth’s centre.
Thinking escape velocity depends on the mass of the escaping object.
Using orbital speed as though it were the same as escape speed.
Forgetting that gravitational potential energy is negative when zero is chosen at infinity.
Assuming a satellite is weightless because gravity is zero.

Quick Revision

F=Gm₁m₂/r²
g=GM/r² outside a spherical mass
g_h=g[R/(R+h)]²
U=−GMm/r
V=−GM/r
v_e=√(2GM/R)=√(2gR)
v_o=√(GM/r)
T=2π√(r³/GM)
For a circular orbit, v_e=√2 v_o at the same radius

Class 11 Physics – Gravitation