Gravitation: NEB Class 11 Physics Guide

One idea across scales

How Gravitation connects Mechanics

Gravitation applies Newton’s laws, energy conservation and circular motion to planets and satellites. Use the CDC Physics Grade 11 page to confirm Nepal course scope.

The OpenStax universal-gravitation chapter emphasises that the force is attractive and acts along the centre-to-centre line. It also states the point-mass/spherical-symmetry condition that students often omit.

Inverse-square interaction

Newton’s law of universal gravitation

For point masses m₁ and m₂ separated by centre-to-centre distance r, F = Gm₁m₂/r². The forces on the two bodies are equal in magnitude and opposite in direction, consistent with Newton’s third law. G is the universal gravitational constant; distinguish it from g, the local gravitational field strength.

Force per unit test mass

Gravitational field and superposition

Field strength at a point is force per unit test mass: g = F/m. For a spherical source M outside it, g = GM/r² directed toward the centre. Unit N kg⁻¹ is equivalent to m s⁻². The field describes the source and location; force on a chosen test mass is F = mg.

Fields add as vectors. Between two masses, determine each field direction before subtracting or adding magnitudes. A zero-field point is not automatically a zero-potential point because potential is scalar.

Location matters

Variation of g with altitude and depth

Outside an ideal spherical Earth, g = GM/r². At height h above the surface, r = R+h. For h much smaller than R, the binomial approximation gives g_h ≈ g₀(1−2h/R). Use the exact expression unless a small-height approximation is requested.

Inside an ideal uniform-density sphere at depth d, only the enclosed mass contributes to the net field, giving g_d = g₀(1−d/R). Real Earth is not uniformly dense, so this is a model. The OpenStax near-Earth discussion separates field variation from apparent weight in accelerating frames.

Scalar energy map

Gravitational potential and potential energy

Taking zero at infinity, gravitational potential of a spherical mass is V = −GM/r and potential energy is U = mV = −GMm/r. The negative sign reflects an attractive bound system: energy must be supplied to separate the masses to infinity.

Near Earth’s surface, ΔU ≈ mgΔh is a local approximation for small height compared with Earth’s radius. Do not use mgh for satellite-scale changes. The OpenStax energy treatment uses K+U and the infinity reference to derive escape conditions.

Continuous free fall

Circular satellites and orbital period

For an ideal circular orbit, gravity supplies the radial force: GMm/r² = mv²/r. Satellite mass cancels and v_orbit = √(GM/r). The period is T = 2πr/v = 2π√(r³/GM), consistent with the circular form of Kepler’s third law.

The OpenStax satellite-orbit section relates speed, kinetic energy, potential energy and total energy. In a circular orbit K = GMm/(2r), U = −GMm/r and total E = −GMm/(2r).

“Weightlessness” in orbit does not mean gravity is absent. Astronaut and spacecraft accelerate together in free fall, so a support-force scale reads approximately zero.

Minimum unbound energy

Escape speed

For the ideal minimum escape condition, the object reaches infinity with zero final speed. Energy conservation gives ½mv_e² − GMm/r = 0, hence v_e = √(2GM/r). It is √2 times the circular-orbit speed at the same radius and independent of projectile mass in the ideal model.

Planetary pattern

Kepler’s laws and Newtonian explanation

1. Planets follow ellipses with the Sun at one focus.
2. The line from Sun to planet sweeps equal areas in equal times.
3. For bodies orbiting the same central mass, T² ∝ a³.

The OpenStax Kepler section connects these empirical laws to gravity. In circular simplification, a becomes orbital radius r, and Newton’s law directly gives T² = 4π²r³/(GM).

Choose the right quantity

Force, field, potential and energy compared

For two source masses, draw each field contribution before vector addition. For potential, add signed scalar values. A location can have zero net field because vector contributions cancel while its potential remains negative. This distinction is a useful conceptual check and prevents treating potential as if it pointed somewhere.

Dimensions offer a second check: G has units N m² kg⁻²; g has N kg⁻¹; potential has J kg⁻¹; potential energy has joules. If the requested quantity and final unit disagree, revisit the first equation rather than patching the unit afterward.

Exam method

A reliable gravitation solution

1. Draw the masses and measure r centre to centre.
2. State whether point-mass or spherical symmetry is assumed.
3. Choose force, field, potential or energy based on the question.
4. Write the symbolic relation before substituting.
5. For multiple sources, add fields as vectors and potentials as scalars.
6. For orbits, equate gravity to required radial force.
7. Check inverse-square scaling, sign, unit and limiting behaviour.

Use the Gravitation study guide for a revision sequence. For online or physical NEB tuition, call 9846662070 with one attempted derivation.

Closed-book checkpoint

Practice tasks

1. Find how force changes when distance triples.
2. Explain G versus g with units.
3. Find g/g₀ at altitude h = R.
4. Explain why potential is negative with zero at infinity.
5. Derive circular orbital speed.
6. Derive escape speed by energy conservation.
7. Explain orbital weightlessness without saying gravity is zero.

References and next steps

Sources and related study guides

• CDC Nepal: Physics Grade 11
• OpenStax: Universal Gravitation
• OpenStax: Gravitation Near Earth
• OpenStax: Gravitational Potential Energy
• OpenStax: Satellite Orbits and Energy
• PhET: Gravity and Orbits

Continue with the Gravitation study guide for retrieval practice and an exam-ready learning sequence. Scope and sources were checked on 2 August 2026. Follow current CDC, NEB and college instructions if requirements change.