Gravitational Fields: Question 1

Worked solution

Step 1: Recall the formula for the field of a point mass

Combining Newton’s law of gravitation $F=GMm/r^2$ with the definition of gravitational field strength $g=F/m$ gives, for a point mass (or a uniform sphere, treated as a point mass at its centre): $g = \frac{GM}{r^2}$

Step 2: Substitute the values

Here $M = 3.20\times10^{20}\text{ kg}$ and $r = 4.00\times10^5\text{ m}$ (the radius, since the question asks for the field at the surface): $g = \frac{(6.67\times10^{-11})(3.20\times10^{20})}{(4.00\times10^5)^2}$

Working the numerator: $6.67\times3.20=21.344$, and $10^{-11}\times10^{20}=10^{9}$, so the numerator is $21.344\times10^{9}=2.1344\times10^{10}$.

Working the denominator: $(4.00\times10^5)^2 = 16.0\times10^{10}=1.60\times10^{11}$.

$g = \frac{2.1344\times10^{10}}{1.60\times10^{11}} = 0.1334\text{ N kg}^{-1}$

Check by recomputing differently: $6.67/1.60=4.169$ and $3.20/16.0=0.200$, so $g=4.169\times0.200\times10^{-11+20-10}=0.8338\times10^{-1}$… regrouping the powers directly instead ($10^{-11+20}\div10^{11}=10^{-2}$) gives $g=(6.67\times3.20/16.0)\times10^{-2}=(21.344/16.0)\times10^{-2}=1.334\times10^{-2}\times10^{1}=0.1334$. Both routes agree: $g \approx 0.133\text{ N kg}^{-1}$ (3 s.f.).

Why the other options are wrong

• A ($0.0334\text{ N kg}^{-1}$): this comes from using the diameter ($8.00\times10^5\text{ m}$) instead of the radius. Since $g\propto1/r^2$, using a value of $r$ twice as large makes $g$ four times too small: $0.1334/4=0.0334$.
• C ($1.33\text{ N kg}^{-1}$): this is the correct value multiplied by $10$. A power-of-ten slip, for example treating $4.00\times10^5$ as if it were $4.00\times10^4$.
• D ($5.34\times10^4\text{ N kg}^{-1}$): this comes from forgetting to square $r$, i.e. calculating $GM/r$ instead of $GM/r^2$, which gives a value with completely the wrong order of magnitude.

Final answer

• The gravitational field strength at the surface of Piri is $g=\boxed{0.133}\text{ N kg}^{-1}$, option B.