Gravitation — Practice

Question 1

State Newton's universal law of gravitation.

Accepted Answer

Step 1 — Write the formal statement.
  Every particle of matter 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.

Step 2 — Symbolic form.
  F = G × m₁ × m₂ / r².
  G = 6.674 × 10⁻¹¹ N·m²/kg², the universal gravitational constant.

✦ Answer: As above. F ∝ m₁m₂ / r².

Question 2

Find the weight of a 60 kg person on (a) Earth and (b) the Moon. (g_Earth = 10 m/s², g_Moon = 1.6 m/s²)

Accepted Answer

Step 1 — W = mg.
  W_Earth = 60 × 10 = 600 N.
  W_Moon = 60 × 1.6 = 96 N.

✦ Answer: (a) 600 N on Earth, (b) 96 N on the Moon.

Note: Mass is the same (60 kg) in both places. Only weight changes because g is different.

Question 3

A box of weight 200 N is placed on a table. The base of the box has area 0.5 m². What pressure does the box exert on the table?

Accepted Answer

Step 1 — Use P = F/A.
  F = weight = 200 N. A = 0.5 m².

Step 2 — Substitute.
  P = 200 / 0.5 = 400 Pa.

✦ Answer: 400 Pa.

Question 4

A ball is dropped from a height of 20 m. Find the time it takes to reach the ground. (g = 10 m/s²)

Accepted Answer

Step 1 — Use h = ut + ½gt² with u = 0.
  20 = 0 + ½ × 10 × t²
  20 = 5t²
  t² = 4
  t = 2 s.

✦ Answer: 2 seconds.

Question 5

Why does an iron nail sink in water but a much heavier ship made of iron floats?

Accepted Answer

Step 1 — Key principle: floating depends on DENSITY (or volume relative to mass), not on weight alone.

Step 2 — Iron nail.
  Nail is solid iron. Density of iron (~ 7800 kg/m³) > density of water (1000 kg/m³). Buoyant force < weight → SINKS.

Step 3 — Iron ship.
  Ship is HOLLOW. Its overall (average) density (mass / total volume including the air pocket) is LESS than water's density. By Archimedes' principle, the ship displaces a huge volume of water; weight of displaced water = ship's total weight → FLOATS.

✦ Answer: A nail's solid-iron density > water's. A ship's hollow design gives a much LOWER average density (mass/volume including the interior air), allowing buoyant force to equal its weight. Hence ship floats; nail sinks.

Question 6

Two bodies of masses 100 kg and 200 kg are placed 10 m apart. Find the gravitational force between them. (G = 6.67 × 10⁻¹¹ N·m²/kg²)

Accepted Answer

Step 1 — Use F = G × m₁m₂ / r².

Step 2 — Substitute.
  F = (6.67 × 10⁻¹¹) × (100 × 200) / (10²)
   = (6.67 × 10⁻¹¹) × 20,000 / 100
   = (6.67 × 10⁻¹¹) × 200
   = 1.334 × 10⁻⁸ N.

✦ Answer: ≈ 1.33 × 10⁻⁸ N (about 0.000000013 N).

That's incredibly small — which is why we don't feel gravity from each other in a room.

Question 7

A stone is thrown vertically upward with initial velocity 20 m/s. Find (a) the maximum height reached, (b) the time to reach the maximum height. (g = 10 m/s²)

Accepted Answer

Step 1 — Take upward as positive. u = +20, a = −10 m/s² (gravity opposes upward motion). At max height, v = 0.

Step 2 (a) — Use v² = u² + 2as.
  0 = (20)² − 2 × 10 × h
  h = 400/20 = 20 m.

Step 3 (b) — Use v = u + at.
  0 = 20 − 10t → t = 2 s.

✦ Answer: (a) Maximum height = 20 m. (b) Time to max height = 2 s.

Question 8

Find the pressure exerted by sea water (density 1030 kg/m³) at a depth of 100 m. (g = 10 m/s²)

Accepted Answer

Step 1 — Use P = ρgh.

Step 2 — Substitute.
  P = 1030 × 10 × 100
   = 1,030,000 Pa
   ≈ 1.03 × 10⁶ Pa
   ≈ 1030 kPa
   ≈ 10 atm (atmospheric pressures).

Step 3 — Note: this is JUST the water pressure at that depth. The total pressure at that depth is water pressure + atmospheric pressure (~1 atm) = ~ 11 atm.

✦ Answer: P ≈ 1.03 × 10⁶ Pa (≈ 10 atm of water pressure).

This is why deep-sea submarines need extremely thick hulls — at 100 m depth, the pressure on each square metre of hull is ~ 10 tons of force.

Question 9

An object has mass 200 g and volume 50 cm³. Find its density and relative density. State whether it will sink or float in water.

Accepted Answer

Step 1 — Density = mass/volume = 200 g / 50 cm³ = 4 g/cm³.
  In SI: 4 g/cm³ = 4000 kg/m³.

Step 2 — Relative density = density of object / density of water.
  = 4000 / 1000 = 4. (Equivalently, 4 g/cm³ / 1 g/cm³ = 4.)

Step 3 — Verdict.
  Density (4000) > density of water (1000) → R.D. > 1 → SINKS.

✦ Answer: Density = 4 g/cm³ = 4000 kg/m³. Relative density = 4. The object will SINK in water.

Question 10

Why is the value of g less at the equator than at the poles?

Accepted Answer

Step 1 — Two effects make g smaller at the equator.

(i) Earth's shape:
  Earth is not a perfect sphere — it bulges at the equator (oblate spheroid). The equatorial radius (R_eq ≈ 6378 km) is LARGER than the polar radius (R_pole ≈ 6357 km).
  Since g = GM/R², a larger R gives smaller g at the equator.

(ii) Earth's rotation:
  The earth spins on its axis. At the equator, this spin creates a 'centrifugal effect' that effectively subtracts from g (this is a non-inertial-frame effect; in the rotating frame of the Earth, you feel slightly less weight at the equator).
  At the poles, you're on the axis of rotation — no centrifugal effect.

Step 2 — Combined: g_pole ≈ 9.83 m/s², g_equator ≈ 9.78 m/s² (about 0.5% difference).

✦ Answer: (i) Earth is oblate — larger radius at the equator → g = GM/R² gives smaller g there. (ii) Earth's rotation produces a centrifugal effect at the equator that effectively reduces g, while the poles have no such effect.

Question 11

A cubical wooden block of side 10 cm and density 600 kg/m³ floats in water. (a) What fraction of the block is submerged? (b) What is the depth of the submerged part below the water surface?

Accepted Answer

Step 1 — Find total volume.
  V_total = (0.1 m)³ = 10⁻³ m³.

Step 2 (a) — Fraction submerged.
  By Archimedes: fraction submerged = ρ_block / ρ_water = 600 / 1000 = 0.6.
  So 60 % of the block is under water.

Step 3 (b) — Depth of the submerged part.
  Submerged volume = 0.6 × V_total = 0.6 × 10⁻³ = 6 × 10⁻⁴ m³.
  Since the cross-section (top view) of the block is 0.1 × 0.1 = 0.01 m², the depth h is given by:
  V_submerged = A × h ⇒ 6 × 10⁻⁴ = 0.01 × h ⇒ h = 0.06 m = 6 cm.

✦ Answer: (a) 60 % submerged; (b) Submerged depth = 6 cm.

Question 12

Calculate the force of gravitational attraction between Earth (mass 6 × 10²⁴ kg) and a person of mass 60 kg standing on Earth's surface. (G = 6.67 × 10⁻¹¹, R_Earth = 6.4 × 10⁶ m). Show that the result is the person's weight on Earth.

Accepted Answer

Step 1 — Use F = G × Mm/R².

Step 2 — Substitute.
  F = (6.67 × 10⁻¹¹) × (6 × 10²⁴ × 60) / (6.4 × 10⁶)²
   = (6.67 × 10⁻¹¹) × (3.6 × 10²⁶) / (4.096 × 10¹³)
   = (6.67 × 10⁻¹¹) × 8.79 × 10¹²
   ≈ 586 N.

Step 3 — Compare to weight from W = mg.
  W = 60 × 9.8 = 588 N. ✓ (within rounding error of 586 N).

Step 4 — Conclusion.
  The two methods give the same answer because g IS defined as GM/R². So mg = m × GM/R² = GMm/R² is the force of gravity — by construction.

✦ Answer: F ≈ 586 N, which equals W = mg = 588 N. This proves that weight = gravitational pull of Earth on the body.

Question 13

A body is thrown vertically upward from a tower of height 25 m with velocity 15 m/s. Find the total time the body remains in the air before hitting the ground. (g = 10 m/s²)

Accepted Answer

Step 1 — Pick up as positive. Body starts at top of tower (set this as origin). u = +15, a = −10. Ground is at h = −25 (below origin).

Step 2 — Use h = ut + ½at².
  −25 = 15t + ½(−10)t²
  −25 = 15t − 5t²
  5t² − 15t − 25 = 0
  t² − 3t − 5 = 0.

Step 3 — Solve quadratic.
  t = (3 ± √(9 + 20))/2 = (3 ± √29)/2.
  Take the positive root: t = (3 + 5.385)/2 ≈ 4.19 s.
  (The negative root would correspond to time before the throw — unphysical.)

✦ Answer: Total time ≈ 4.19 s.

Sanity check: time to max height = u/g = 15/10 = 1.5 s. So body goes up for 1.5 s, then falls. From max height, body falls (25 + ½ × 10 × 1.5²) = 25 + 11.25 = 36.25 m. Fall time: √(2 × 36.25/10) = √7.25 ≈ 2.69 s. Total = 1.5 + 2.69 ≈ 4.19 s. ✓

Question 14

A ship floats with 5 m of its hull below the water surface in fresh water. When it sails into sea water, the hull is now only 4.85 m below the surface. Why? (density_seawater = 1030 kg/m³, density_freshwater = 1000 kg/m³)

Accepted Answer

Step 1 — Use Archimedes: weight of ship = weight of displaced water.
  In fresh water: W = ρ_fw × V_fw × g, where V_fw = submerged volume in fresh water.
  In sea water: W = ρ_sw × V_sw × g.

Step 2 — Same ship → same weight in both cases.
  ρ_fw × V_fw = ρ_sw × V_sw
  1000 × V_fw = 1030 × V_sw
  V_sw / V_fw = 1000/1030 ≈ 0.971.

Step 3 — Volumes scale linearly with depth (constant cross-section).
  Depth in sea water = 0.971 × Depth in fresh water = 0.971 × 5 ≈ 4.85 m. ✓

Step 4 — Physical interpretation.
  Sea water is DENSER than fresh water. Same displaced volume in sea water has more mass → more upward buoyant force per unit volume. So the ship needs to displace LESS volume to balance its weight → sits HIGHER (less depth submerged) in sea water.

✦ Answer: Sea water (denser than fresh water) gives more buoyant force per unit volume of displacement. The ship needs to displace less volume to support its weight → sits ~3% higher in sea water. Hence depth submerged decreases from 5.00 m to 4.85 m.

Question 15

Why is it easier to swim in sea water than in fresh water?

Accepted Answer

Step 1 — Density comparison.
  Sea water density ≈ 1030 kg/m³. Fresh water density ≈ 1000 kg/m³. Sea water is ~3% denser.

Step 2 — Apply Archimedes' principle.
  Buoyant force on a swimmer = ρ_water × V_submerged × g.
  Greater density → greater buoyant force for the same submerged volume.
  Sea water gives ~3% more buoyant force than fresh water for the same swimmer.

Step 3 — What this means for the swimmer.
  The swimmer's apparent weight (actual weight − buoyant force) is reduced more in sea water.
  Less downward 'effective weight' means less energy is needed to stay afloat → easier to swim.

Step 4 — Extreme example.
  The Dead Sea (Israel/Jordan) has water density ≈ 1240 kg/m³ (extremely salty). Buoyant force is so high that people FLOAT on the surface with much of their body above water — they can't even fully submerge themselves without effort.

✦ Answer: Sea water is denser than fresh water → larger buoyant force on the swimmer's body → smaller effective weight → swimmer doesn't have to expend as much energy to stay afloat. Easiest case: Dead Sea, where people automatically float without swimming.

Practice — Gravitation

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