2018 A/L Physics Past Paper · Question 11 (English Medium)

Answer: (1) $R_A = \sqrt{2}R_B$

What the question asks

We know: g on the surface of planet A equals g on the surface of planet B.

The mass of A is twice the mass of B: $M_A = 2M_B$.

We want: how is $R_A$ related to $R_B$?

Relevant theory

The intensity of a gravitational field at such a point is the force acting on a unit mass placed at that point.

Required equations

$g = \frac{GM}{R^2}$

Reasoning and calculation

What stays the same? $G$, and $g$ (it is equal on both planets).

What changes? The mass $M$ (it doubles) and the radius $R$.

For $g$ to stay the same, $GM/R^2$ must stay the same.

So $R^2$ must grow with $M$.

Twice the mass needs twice $R^2$.

$\frac{GM_A}{R_A^2} = \frac{GM_B}{R_B^2}$

$\frac{R_A^2}{R_B^2} = \frac{M_A}{M_B} = 2$

$1.414\,R_B$

$R_A = \sqrt{2}\,R_B$

Reading the options

$$\frac{R_A}{R_B} = \sqrt{2} \approx 1.41$$

The options give these ratios: $\sqrt{2}$, $2$, $1/\sqrt{2}$, $1/2$ and $1$.

Option (1) matches.

Checks

$sqrt(MA / MB)$

Does it make sense?

A has more mass, so it pulls harder.

To pull the same, A must be bigger. So $R_A$ is larger than $R_B$.

But not twice as large, because $R$ is squared.

Conclusion

$R_A = \sqrt{2}R_B$

$$R_A^2 = 2R_B^2$$

$$R_A = \sqrt{2}\,R_B$$

The answer is option (1).

R ∝ M

When g is equal, $R^2$ grows with $M$. So $R$ grows with $\sqrt{M}$.