D.1 Gravitational fields

Newton's law of gravitation gives $F=G\frac{M_1{M}_2}{r^2}$. The masses are identical in both scenarios, so only the distance changes.

First scenario: $F_1=G\frac{m\cdot m}{d^2}=\frac{G{m}^2}{d^2}$. Second scenario: $F_2=G\frac{m\cdot m}{(3d{)}^2}=\frac{G{m}^2}{9d^2}$.

$\frac{F_1}{F_2}=\frac{G{m}^2/d^2}{G{m}^2/9d^2}=9$

Tripling the distance reduces the force by a factor of $3^2=9$, so $F_1:F_2=9$.

We need the ratio $F_1/F_2$ when the separation increases by a factor of 3. Since $F\propto 1/r^2$, multiplying $r$ by 3 divides $F$ by $3^2$.

The option 3 would be correct if the force followed a simple inverse relationship ($F\propto 1/r$), but gravity follows an inverse square law, so this is incorrect.

The option 6 has no physical basis in Newton's law and does not correspond to any standard power law relationship here.

The option 27 corresponds to $3^3$, which would apply to a hypothetical inverse-cube law, not the inverse square law of gravity.

The correct ratio is 9, because $F\propto 1/r^2$ and the distance is tripled: $3^2=9$.

Planet mass $M=4.0\times 10^{26}$ kg, spacecraft mass $m=2000$ kg, distance $r=1.5\times 10^{10}$ m, $G=6.67\times 10^{-11}$ N m² kg⁻².

$F=G\frac{Mm}{r^2}=\frac{(6.67\times 10^{-11})(4.0\times 10^{26})(2000)}{(1.5\times 10^{10}{)}^2}$

Numerator: $(6.67\times 10^{-11})(4.0\times 10^{26})(2000)=5.336\times 10^{19}$. Denominator: $(1.5\times 10^{10}{)}^2=2.25\times 10^{20}$.

$F=\frac{5.336\times 10^{19}}{2.25\times 10^{20}}\approx 2.4\times 10^9\text{ N}$

The gravitational force is approximately $2.4\times 10^9$ N.

With $G\approx 7\times 10^{-11}$, $M\approx 4\times 10^{26}$, $m=2\times 10^3$, $r^2\approx 2.25\times 10^{20}$, the numerator is $\sim 5\times 10^{19}$ and the result is $\sim 10^9$ N.

This is two orders of magnitude too small. It would result from a significant arithmetic error such as forgetting to square $r$ properly or using incorrect mass values.

This is roughly five times smaller than expected. It does not match a careful substitution of the given values.

This value is half the correct answer and would arise from an error such as not doubling the spacecraft mass or using $r$ instead of $r^2$ partially.

A careful calculation confirms $2.4\times 10^9$ N as the correct gravitational force.