2021 A/L Physics Past Paper · Question 20 (English Medium)

Answer: (1) $27.5\ \text{N kg}^{-1}$

What the question asks

Find the gravitational intensity $g_J$ at the surface of Jupiter.

We only need how $g$ changes when the size and the density change.

Relevant theory

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

Hence it is seen that the acceleration due to gravity and the gravitational field intensity on the surface of Earth take the same value.

The mass of a unit volume of a substance is termed its density.

Required equations

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

$M = \rho \times \tfrac{4}{3}\pi R^3$

$g = \tfrac{4}{3}\pi G \rho R$

Reasoning and calculation

What stays the same: $G$ and $\tfrac{4}{3}\pi$.

What changes: the density $\rho$ and the radius $R$.

The diameter is $11$ times, so the radius is $11$ times too.

So the gravitational intensity is proportional to the density times the radius:

$$g \propto \rho R$$

Take the ratio of Jupiter to Earth:

$$\dfrac{g_J}{g_E} = \dfrac{\rho_J R_J}{\rho_E R_E} = \dfrac{1}{4} \times 11$$

Simplify:

$$\dfrac{g_J}{g_E} = \dfrac{11}{4} = 2.75$$

$$g_J = 2.75 \times 10 = 27.5\ \text{N kg}^{-1}$$

Reading the options

The value $27.5\ \text{N kg}^{-1}$ is option (1).

The options $110$ and $440$ are too big. They ignore the density, or turn it upside down.

Checks

Dimension check:

$[G\rho R] = M^{-1}L^3T^{-2} \times ML^{-3} \times L = LT^{-2}$

This is the dimension of $g$ ($\text{N kg}^{-1} = \text{m s}^{-2}$). It matches.

$g * d * D$

Does it make sense?

Jupiter is much bigger than Earth, so a bigger $g$ is right.

But the density is smaller, so $g$ is not $11$ times bigger. $2.75$ is between $1$ and $11$.

A limiting case. If the density were the same as Earth, we would get $\dfrac{g_J}{g_E} = 11$.

With a quarter of the density the factor is a quarter of that: $\dfrac{11}{4}$.

Conclusion

$g_J = \dfrac{1}{4} \times 11 \times 10 = 27.5\ \text{N kg}^{-1}$

$g \propto \rho R$, so $\dfrac{g_J}{g_E} = \dfrac{1}{4} \times 11 = 2.75$.

$g_J = 27.5\ \text{N kg}^{-1}$. The answer is (1).

For a planet, $g$ goes with density times radius.

Write the ratio of the two planets and the constants cancel.

Exam tip