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9702 · 13.3

Gravitational field of a point mass — practice questions

Practice and worked examples for 9702 Gravitational field of a point mass. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Calculate the gravitational field strength on the surface of Mars, given its mass is 6.42×1023 kg6.42 \times 10^{23} \text{ kg} and its radius is 3.39×106 m3.39 \times 10^6 \text{ m}. Use the universal gravitational constant 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 known values:
    • Mass of Mars (MM) =6.42×1023 kg= 6.42 \times 10^{23} \text{ kg}
    • Radius of Mars (rr) =3.39×106 m= 3.39 \times 10^6 \text{ m}
    • Universal Gravitational Constant (GG) =6.67×10−11 N m2kg−2= 6.67 \times 10^{-11} \text{ N m}^2 \text{kg}^{-2}
  2. Recall the formula for gravitational field strength: g=GMr2g = \frac{GM}{r^2}
  3. Substitute the values into the formula: g=(6.67×10−11)×(6.42×1023)(3.39×106)2g = \frac{(6.67 \times 10^{-11}) \times (6.42 \times 10^{23})}{(3.39 \times 10^6)^2}
  4. Calculate the denominator (radius squared): (3.39×106)2=1.14921×1013 m2(3.39 \times 10^6)^2 = 1.14921 \times 10^{13} \text{ m}^2
  5. Calculate the numerator (GMGM): (6.67×10−11)×(6.42×1023)=4.28114×1013 N m2kg−1(6.67 \times 10^{-11}) \times (6.42 \times 10^{23}) = 4.28114 \times 10^{13} \text{ N m}^2 \text{kg}^{-1}
  6. Divide the numerator by the denominator: g=4.28114×10131.14921×1013≈3.725 N kg−1g = \frac{4.28114 \times 10^{13}}{1.14921 \times 10^{13}} \approx 3.725 \text{ N kg}^{-1}
  7. Round to an appropriate number of significant figures (e.g., 3 s.f.): g≈3.73 N kg−1g \approx 3.73 \text{ N kg}^{-1}

Worked example 2

A satellite of mass 1200 kg is in a stable circular orbit at an altitude of 500 km above the Earth's surface. Calculate the work done required to move it to a higher stable orbit at an altitude of 2000 km. (Mass of Earth, M = 5.97×10245.97 \times 10^{24} kg; Radius of Earth, RE=6.37×106R_E = 6.37 \times 10^6 m; G=6.67×10−11G = 6.67 \times 10^{-11} N m2^2 kg−2^{-2})

Show solution outline
  1. Identify the principle: The work done is the change in the satellite's gravitational potential energy (GPE). Work Done = Ep,final−Ep,initialE_{p, final} - E_{p, initial}
  2. Calculate initial and final orbital radii: Remember to add the Earth's radius to the altitude.
    • Initial altitude h1=500 km=5.00×105 mh_1 = 500 \text{ km} = 5.00 \times 10^5 \text{ m}
    • Initial radius r1=RE+h1=(6.37×106)+(0.50×106)=6.87×106 mr_1 = R_E + h_1 = (6.37 \times 10^6) + (0.50 \times 10^6) = 6.87 \times 10^6 \text{ m}
    • Final altitude h2=2000 km=2.00×106 mh_2 = 2000 \text{ km} = 2.00 \times 10^6 \text{ m}
    • Final radius r2=RE+h2=(6.37×106)+(2.00×106)=8.37×106 mr_2 = R_E + h_2 = (6.37 \times 10^6) + (2.00 \times 10^6) = 8.37 \times 10^6 \text{ m}
  3. Recall the GPE formula: Ep=−GMmrE_p = -\frac{GMm}{r}
  4. Calculate initial GPE (Ep1E_{p1}): Ep1=−(6.67×10−11)×(5.97×1024)×12006.87×106E_{p1} = -\frac{(6.67 \times 10^{-11}) \times (5.97 \times 10^{24}) \times 1200}{6.87 \times 10^6} Ep1=−4.778×10176.87×106=−6.955×1010 JE_{p1} = -\frac{4.778 \times 10^{17}}{6.87 \times 10^6} = -6.955 \times 10^{10} \text{ J}
  5. Calculate final GPE (Ep2E_{p2}): Ep2=−(6.67×10−11)×(5.97×1024)×12008.37×106E_{p2} = -\frac{(6.67 \times 10^{-11}) \times (5.97 \times 10^{24}) \times 1200}{8.37 \times 10^6} Ep2=−4.778×10178.37×106=−5.708×1010 JE_{p2} = -\frac{4.778 \times 10^{17}}{8.37 \times 10^6} = -5.708 \times 10^{10} \text{ J}
  6. Calculate the work done (change in GPE): Work Done = Ep2−Ep1=(−5.708×1010)−(−6.955×1010)E_{p2} - E_{p1} = (-5.708 \times 10^{10}) - (-6.955 \times 10^{10}) Work Done = 1.247×1010 J1.247 \times 10^{10} \text{ J}
  7. Final Answer: The work done required is 1.25×10101.25 \times 10^{10} J. The positive value indicates energy must be supplied to the satellite.
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