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Gravitational Fields: Question 8

Syllabus 13.4

Multiple choice A2 1 mark

Two point masses, C and D, are fixed in space. Mass C is 3.00×1023 kg3.00\times10^{23}\text{ kg} and mass D is 7.00×1023 kg7.00\times10^{23}\text{ kg}. The distance between C and D is 5.00×107 m5.00\times10^7\text{ m}.

Take G=6.67×10−11 N m2 kg−2G = 6.67\times10^{-11}\text{ N m}^2\text{ kg}^{-2}.

What is the gravitational potential at the midpoint of the line joining C and D (a distance of 2.50×107 m2.50\times10^7\text{ m} from each mass)?

Choose an answer to check it, then compare with the worked solution below.
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Worked solution

Step 1: Recall that gravitational potential is a scalar

Unlike gravitational field strength, gravitational potential ϕ\phi has no direction. It is a scalar. The total potential at a point due to several masses is simply the ordinary (algebraic) sum of the potential due to each mass on its own: ϕtotal=ϕC+ϕD\phi_{\text{total}} = \phi_C + \phi_D

Step 2: Calculate the potential due to each mass at the midpoint

Both C and D are r=2.50×107 mr=2.50\times10^7\text{ m} from the midpoint. Using ϕ=−GM/r\phi=-GM/r for each:

For C (MC=3.00×1023 kgM_C=3.00\times10^{23}\text{ kg}): ϕC=−(6.67×10−11)(3.00×1023)2.50×107\phi_C = -\frac{(6.67\times10^{-11})(3.00\times10^{23})}{2.50\times10^7}

Numerator: 6.67×3.00=20.016.67\times3.00=20.01, and 10−11×1023=101210^{-11}\times10^{23}=10^{12}, so the numerator is 2.001×10132.001\times10^{13}. ϕC=−2.001×10132.50×107=−8.004×105 J kg−1\phi_C = -\frac{2.001\times10^{13}}{2.50\times10^7} = -8.004\times10^5\text{ J kg}^{-1}

For D (MD=7.00×1023 kgM_D=7.00\times10^{23}\text{ kg}): ϕD=−(6.67×10−11)(7.00×1023)2.50×107\phi_D = -\frac{(6.67\times10^{-11})(7.00\times10^{23})}{2.50\times10^7}

Numerator: 6.67×7.00=46.696.67\times7.00=46.69, and 10−11×1023=101210^{-11}\times10^{23}=10^{12}, so the numerator is 4.669×10134.669\times10^{13}. ϕD=−4.669×10132.50×107=−1.8676×106 J kg−1\phi_D = -\frac{4.669\times10^{13}}{2.50\times10^7} = -1.8676\times10^6\text{ J kg}^{-1}

Step 3: Add the two potentials

ϕtotal=ϕC+ϕD=(−8.004×105)+(−1.8676×106)=−2.6680×106 J kg−1≈−2.67×106 J kg−1\phi_{\text{total}} = \phi_C+\phi_D = (-8.004\times10^5)+(-1.8676\times10^6) = -2.6680\times10^6\text{ J kg}^{-1} \approx -2.67\times10^6\text{ J kg}^{-1}

Check by combining the masses first (valid since both are the same distance rr from the midpoint): ϕtotal=−G(MC+MD)/r=−(6.67×10−11)(1.00×1024)/(2.50×107)=−(6.67×1013)/(2.50×107)=−2.668×106 J kg−1\phi_{\text{total}}=-G(M_C+M_D)/r = -(6.67\times10^{-11})(1.00\times10^{24})/(2.50\times10^7) = -(6.67\times10^{13})/(2.50\times10^7) = -2.668\times10^6\text{ J kg}^{-1}, matching (to rounding).

Why the other options are wrong

  • A (0 J kg−10\text{ J kg}^{-1}): this comes from wrongly treating potential as a vector and assuming the contributions cancel, as they would for a field-strength (vector) calculation, but potential is a scalar, so both negative contributions add rather than cancel.
  • B (−8.00×105 J kg−1-8.00\times10^5\text{ J kg}^{-1}): this is ϕC\phi_C alone, forgetting to add the contribution from D.
  • D (+2.67×106 J kg−1+2.67\times10^6\text{ J kg}^{-1}): this has the correct magnitude but the wrong sign, from forgetting the minus sign in ϕ=−GM/r\phi=-GM/r.

Final answer

  • The gravitational potential at the midpoint is ϕ=−2.67×106 J kg−1\phi=\boxed{-2.67\times10^6}\text{ J kg}^{-1}, option C.