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9702 · 13.2

Gravitational force between point masses — practice questions

Practice and worked examples for 9702 Gravitational force between point masses. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Calculate the gravitational force between Earth (mass 5.97×10245.97 \times 10^{24} kg) and the Moon (mass 7.35×10227.35 \times 10^{22} kg) if their average centre-to-centre distance is 3.84×1083.84 \times 10^8 m. Use G=6.67×10−11 N m2kg−2G = 6.67 \times 10^{-11} \text{ N m}^2 \text{kg}^{-2}.

Show solution outline
  1. Identify given values: m1=5.97×1024m_1 = 5.97 \times 10^{24} kg (Earth's mass) m2=7.35×1022m_2 = 7.35 \times 10^{22} kg (Moon's mass) r=3.84×108r = 3.84 \times 10^8 m (centre-to-centre distance) G=6.67×10−11 N m2kg−2G = 6.67 \times 10^{-11} \text{ N m}^2 \text{kg}^{-2} (universal gravitational constant)
  2. Write down the formula: F=Gm1m2r2F = G \frac{m_1 m_2}{r^2}
  3. Substitute the values into the formula: F=(6.67×10−11)(5.97×1024)(7.35×1022)(3.84×108)2F = (6.67 \times 10^{-11}) \frac{(5.97 \times 10^{24})(7.35 \times 10^{22})}{(3.84 \times 10^8)^2}
  4. Calculate the product of masses: (5.97×1024)×(7.35×1022)=4.38795×1047(5.97 \times 10^{24}) \times (7.35 \times 10^{22}) = 4.38795 \times 10^{47}
  5. Calculate the square of the distance: (3.84×108)2=1.47456×1017(3.84 \times 10^8)^2 = 1.47456 \times 10^{17}
  6. Perform the division: 4.38795×10471.47456×1017≈2.9758×1030\frac{4.38795 \times 10^{47}}{1.47456 \times 10^{17}} \approx 2.9758 \times 10^{30}
  7. Multiply by G: F=(6.67×10−11)×(2.9758×1030)≈1.9848×1020F = (6.67 \times 10^{-11}) \times (2.9758 \times 10^{30}) \approx 1.9848 \times 10^{20} N
  8. State the final answer (to 3 significant figures): The gravitational force between Earth and the Moon is approximately 1.98×10201.98 \times 10^{20} N.

Worked example 2

Two large spheres with masses MA=200M_A = 200 kg and MB=500M_B = 500 kg are placed with their centres 4.0 m apart. A smaller object with mass m=10m = 10 kg is placed on the line connecting their centres, at a distance of 1.0 m from the 200 kg sphere. Calculate the net gravitational force on the 10 kg object. Use G=6.67×10−11 N m2kg−2G = 6.67 \times 10^{-11} \text{ N m}^2 \text{kg}^{-2}.

Show solution outline
  1. Identify forces and distances: The 10 kg mass (m) is attracted by both MAM_A and MBM_B. Let FAF_A be the force from MAM_A and FBF_B be the force from MBM_B. These forces act in opposite directions along the line connecting the centres.
    • Distance to MAM_A, rA=1.0r_A = 1.0 m.
    • Distance to MBM_B, rB=4.0 m−1.0 m=3.0r_B = 4.0 \text{ m} - 1.0 \text{ m} = 3.0 m.
  2. Calculate the force from MAM_A (FAF_A): FA=GMAmrA2=(6.67×10−11)(200)(10)(1.0)2F_A = G \frac{M_A m}{r_A^2} = (6.67 \times 10^{-11}) \frac{(200)(10)}{(1.0)^2} FA=(6.67×10−11)×2000=1.334×10−7F_A = (6.67 \times 10^{-11}) \times 2000 = 1.334 \times 10^{-7} N (directed towards MAM_A).
  3. Calculate the force from MBM_B (FBF_B): FB=GMBmrB2=(6.67×10−11)(500)(10)(3.0)2F_B = G \frac{M_B m}{r_B^2} = (6.67 \times 10^{-11}) \frac{(500)(10)}{(3.0)^2} FB=(6.67×10−11)50009=3.7056×10−8F_B = (6.67 \times 10^{-11}) \frac{5000}{9} = 3.7056 \times 10^{-8} N (directed towards MBM_B).
  4. Calculate the net force (FnetF_{net}): Since the forces are in opposite directions, we find the net force by subtracting the smaller force from the larger one. Let's define the direction towards MAM_A as positive. Fnet=FA−FBF_{net} = F_A - F_B Fnet=(1.334×10−7)−(3.7056×10−8)F_{net} = (1.334 \times 10^{-7}) - (3.7056 \times 10^{-8}) To subtract, we can express FBF_B as 0.37056×10−70.37056 \times 10^{-7} N. Fnet=(1.334−0.37056)×10−7=0.96344×10−7F_{net} = (1.334 - 0.37056) \times 10^{-7} = 0.96344 \times 10^{-7} N. Fnet=9.6344×10−8F_{net} = 9.6344 \times 10^{-8} N.
  5. State the final answer: The net gravitational force on the 10 kg object is 9.6×10−89.6 \times 10^{-8} N (to 2 s.f.), directed towards the 200 kg sphere (since FAF_A is larger).
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