NEB Class 11 • Physics • Worked Practice

Gravitation Practice Set: NEB 11 Physics

Move from inverse-square scaling to field, potential, satellites and escape energy. Each worked solution makes the model, direction, unit and physical check visible.

Twenty graded questions with complete reasoning
Force, field, potential, orbit and escape levels
Timed challenge, error diagnosis and FAQs

Gravity supplies the inward acceleration of an ideal orbit.

Practice boundary

What this Gravitation set measures

The problems cover Newton’s law, inverse-square ratios, gravitational field, potential and energy, variation of g, circular satellite motion, escape speed and Kepler scaling. Review the concept guide or study guide before attempting unfamiliar levels. The CDC Physics Grade 11 page is the Nepal scope checkpoint.

The force and symmetry assumptions follow the OpenStax universal-gravitation reference; satellite and energy equations follow its orbital treatment.

Level 1

Universal force and inverse-square scaling

1. Two point masses

Masses 4.0 kg and 6.0 kg are 0.30 m apart. Find attraction using G = 6.67×10⁻¹¹ N m² kg⁻².

Solution: F = Gm₁m₂/r² = 6.67×10⁻¹¹×24/0.09 ≈ 1.78×10⁻⁸ N along the centre line. The forces on the two masses have equal magnitude and opposite direction.

2. Scaling without numbers

One mass becomes three times larger and separation doubles.

Solution: F′/F = 3/2² = 3/4. Force becomes 75% of the original.

3. Centre distance

Spherical bodies of radii 0.20 m and 0.30 m have a surface gap 0.50 m.

Solution: Centre distance is 0.20+0.50+0.30 = 1.00 m. Using 0.50 m would make force four times too large.

4. Find an unknown mass

At r = 2.0 m, m₁ = 10 kg and attraction is 3.335×10⁻¹⁰ N.

Solution: m₂ = Fr²/(Gm₁) = (3.335×10⁻¹⁰×4)/(6.67×10⁻¹¹×10) = 2.0 kg.

Level 2

Field strength and variation of g

5. Field from a spherical source

A planet has mass 6.0×10²⁴ kg and radius 6.4×10⁶ m. Estimate surface g.

Solution: g = GM/R² ≈ (6.67×10⁻¹¹×6.0×10²⁴)/(6.4×10⁶)² ≈ 9.77 m s⁻².

6. Altitude equal to radius

Find field relative to surface at h = R.

Solution: Centre distance is R+h = 2R, so g = g₀(R/2R)² = g₀/4.

7. Small-altitude estimate

For h/R = 0.01, use g ≈ g₀(1−2h/R).

Solution: g ≈ 0.98g₀, a 2% reduction. State that the relation is an approximation for h ≪ R.

8. Depth model

Inside an ideal uniform Earth, find g at depth d = R/3.

Solution: g_d = g₀(1−d/R) = 2g₀/3. Uniform density is a stated model, not a description of real Earth.

9. Force on a test mass

A 3.0 kg object is at a location with g = 4.0 N kg⁻¹.

Solution: F = mg = 12 N toward the source. The field belongs to the location; the force depends on test mass.

Level 3

Potential and energy

10. Potential at radius

Write potential due to mass M at r and compare with 2r.

Solution: V(r) = −GM/r and V(2r) = −GM/(2r). Its magnitude halves; it becomes less negative farther away.

11. Potential-energy change

Move mass m from r to 3r.

Solution: ΔU = −GMm/(3r) − [−GMm/r] = +2GMm/(3r). Positive energy must be supplied in the ideal slow transfer.

12. Near-surface comparison

Why can ΔU ≈ mgh be used for a small 20 m rise?

Solution: When h ≪ R, g changes negligibly and the exact potential difference reduces to the local constant-g approximation. Do not extend mgh to satellite-scale height.

13. Two-source midpoint

At the midpoint between equal masses, compare field and potential.

Solution: Equal opposite field vectors cancel, so net field is zero. Potentials are both negative scalars and add, so potential is not zero.

Level 4

Circular satellites

14. Derive orbit speed

Solution: Set GMm/r² = mv²/r. Cancelling m and one r gives v = √(GM/r). Satellite mass does not appear.

15. Radius scaling

Orbital radius becomes nine times larger around the same planet.

Solution: v ∝ r^−1/2, so speed becomes one third. T ∝ r^3/2, so period becomes 27 times larger.

16. Circular-orbit energy

Given radius r, write K, U and total E.

Solution: K = GMm/(2r), U = −GMm/r and E = −GMm/(2r). Total energy is negative for the bound orbit.

17. Weightlessness

Explain a zero scale reading inside a satellite.

Solution: The satellite and person share gravitational free-fall acceleration, so the scale need not provide a normal support force. Gravity is not zero.

Level 5

Escape speed and mixed energy

18. Derive escape speed

For minimum ideal escape, final speed and potential at infinity are zero.

Solution: ½mv²−GMm/r = 0, hence v_e = √(2GM/r). Projectile mass cancels.

19. Compare orbit and escape

At the same radius, compare v_e with v_orbit.

Solution: v_e/v_orbit = √2. Escape requires greater initial energy than a circular orbit.

20. Escape at larger radius

If launch radius becomes four times larger, compare ideal escape speed.

Solution: v_e ∝ r^−1/2, so it becomes one half.

20-minute mixed challenge

Questions without method labels

Force F acts at r. Find force at 5r with masses unchanged.
At what altitude is ideal field g₀/9?
Compare potential and field at the midpoint of equal masses.
Derive T² ∝ r³ for circular satellites.
A planet has surface circular speed 7 km s⁻¹. Find ideal escape speed.
Explain why doubling satellite mass does not change ideal orbit speed.

Checks: (1) F/25; (2) r = 3R, so h = 2R; (5) 7√2 ≈ 9.9 km s⁻¹.

For a predict–observe–explain extension, use PhET Gravity and Orbits after writing force and velocity predictions.

Error analysis

Common Gravitation failures

Geometry

Use centre distance, not height or surface gap by itself.

Quantity

Field is vector N kg⁻¹; potential is scalar J kg⁻¹.

Approximation

State h ≪ R before using a small-height expression.

Interpretation

Orbital weightlessness is shared free fall, not absence of gravity.

For online or physical NEB tuition, call 9846662070 with your attempted set and first wrong decision.

Frequently asked questions

Questions about the Gravitation practice set

How should I score a Gravitation solution?

Credit the geometry and assumptions, quantity choice, symbolic equation, direction or sign, substitution, unit and scaling check.

Why does satellite mass cancel?

Both gravitational force GMm/r² and required radial term mv²/r are proportional to satellite mass.

Can field be zero while potential is not zero?

Yes. Vector fields can cancel while negative scalar potentials add, as at the midpoint of equal source masses.

When is mgh acceptable?

Use it for small near-surface height changes where g is approximately constant, not for satellite-scale separation.

What score shows readiness?

Aim for at least 80% on a fresh mixed set with no repeated centre-distance, sign or quantity-selection error.

Where can I get NEB Gravitation help?

For current online or physical tuition options, call 9846662070 and confirm timetable, class mode, teacher availability and fees.

References and next steps

Sources and related study guides

Continue with Elasticity to study how materials deform under force. Scope and sources were checked on 2 August 2026; follow current CDC, NEB and college instructions if requirements change.

Level 6

Superposition and multi-stage comparisons

21. Zero-field point between unequal masses

Masses M and 4M are separated by distance d. Let the zero-field point be x from M between them. Opposing field magnitudes satisfy GM/x² = G(4M)/(d−x)². Taking positive square roots gives 1/x = 2/(d−x), so d−x = 2x and x = d/3. The point lies closer to the smaller mass, as expected.

22. Work required to change a circular orbit

A satellite moves from circular radius r to 2r around the same mass. Total circular-orbit energy changes from −GMm/(2r) to −GMm/(4r). The required ideal energy input is +GMm/(4r). The higher orbit has less negative total energy even though its speed is lower.

23. Planet comparison

Planet B has four times Planet A’s mass and twice its radius. Surface field ratio is g_B/g_A = 4/2² = 1, so surface fields are equal. Escape-speed ratio is √[(M_B/R_B)/(M_A/R_A)] = √2.

These questions combine models. Before algebra, write which quantity is vector, which energy reference is used and what remains fixed. A plausible-looking number without those boundaries gives little conceptual evidence.

Final marking check: End each solution with one ratio, unit or limiting-case test. Explain whether increasing source mass or separation should raise or lower the requested quantity before accepting the calculator result.

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