2020 A/L Physics Past Paper · Question 10 (English Medium)

Answer: (3) $\dfrac{15}{2}$ N

What the question asks

We must find: the weight of the $1.0$ kg instrument on the planet's surface, $W = mg_P$; $g_P$ can be compared with Earth's $g = 10\ \text{N kg}^{-1}$.

Relevant theory

The strength of a gravitational field at any point in it is measured by the quantity known as the " Gravitational field intensity " at that point. The intensity of a gravitational field at such a point is the force acting on a unit mass placed at that point.

P is a point situated at a distance r (r > R) of a spherical body of mass M and radius R. In order to determine the gravitational intensity at P, let us place a mass element m at P and find the force on it.

Required equations

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

$W = mg$

Reasoning and calculation

What changes: mass $M \to 3M$, radius $R \to 2R$.

What stays the same: $G$, the instrument's mass $m = 1.0$ kg.

Proportion: $g \propto \dfrac{M}{R^2}$, so $W \propto \dfrac{M}{R^2}$.

$\frac{g_P}{g_E} = \frac{3M/(2R)^2}{M/R^2}$

$\frac{g_P}{g_E} = \frac{3}{4}$

$7.5$

$g_P = 10 \times \frac{3}{4}$

$\frac{15}{2}$

$W = 1.0 \times 7.5$

Reading the options

This gives $\dfrac{15}{2}$ N — option (3).

Checks

Dimension check: the ratio $\dfrac{g_P}{g_E}$ is dimensionless; $W = m \times g$ has unit $\text{kg} \times \text{N kg}^{-1} = \text{N}$ ✓.

$m * g * kM / (kR * kR)$

Second way (step by step): mass alone trebling gives $30$ N; then doubling the radius doubles the distance, which cuts the force to $\dfrac{1}{4}$: $30 \div 4 = 7.5$ N ✓.

Conclusion

$W = 1.0 \times 10 \times \frac{3}{4} = \frac{15}{2}\ \text{N}$

$g_P = 10 \times \dfrac{3}{4} = 7.5\ \text{N kg}^{-1}$; weight $= \dfrac{15}{2}$ N — the answer is (3).

In a question giving "… times the Earth's …", take the ratio of $g = \dfrac{GM}{R^2}$ directly — you never need $G$ or Earth's actual values.

Exam tip