The lines in Figure 4 show the shape of the gravitational field around two stars G and H - AQA - A-Level Physics - Question 4 - 2022 - Paper 2

Annotate Figure 4 by drawing arrows at points X and Y to indicate the direction of the gravitational field. The arrows should point away from G and towards H, showing the direction in which a small mass would accelerate if placed in the field.

To find the radius R of the spherical asteroid P, we use the formula for gravitational field strength at the surface:

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

Here, $g = 0.40 \text{ N kg}^{-1}$ and $M = 2.0 \times 10^{20} \text{ kg}$. Rearranging the formula gives:

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

Substituting the values:

$R = \frac{(6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2)(2.0 \times 10^{20} \text{ kg})}{0.4 \text{ N kg}^{-1}}$

Calculating the above yields:

$R = 1.8 \times 10^8 \text{ m}$.

On Figure 5, sketch a curve that starts at the point corresponding to the surface which is at g = 0.40 N kg⁻¹ when r = R, and then decreases towards 0 as r increases, reflecting the inverse square law of gravitational attraction. The variation should be smooth and decrease continuously as r increases.

The area under the graph between r = R and r = 2R represents the work done per unit mass on an object moving from the surface of P (at r = R) to a distance of 2R. This energy is transferred to the object, changing its gravitational potential as it moves away from the large mass.

To find the acceleration of P, we first calculate the net gravitational force acting on P.

From G, the force exerted on P is calculated as: $F_g = \frac{GMm}{d^2}$ where $M$ is the mass of G, $m$ is the mass of P, and $d$ is the distance HP. Given:

$F = 6.38 \times 10^{12} \text{ N}$

Using Newton's second law, we can calculate acceleration: $a = \frac{F}{m} = \frac{6.38 \times 10^{12}}{2.0 \times 10^{20}} = 3.19 \times 10^{-8} \text{ m s}^{-2}$.

Asteroid P cannot have a circular orbit around H because the gravitational force acting on it from H is insufficient to provide the necessary centripetal force required for a stable circular orbit. The distance HP may cause P's velocity to not be sufficient, leading to instability, and thus it would either drift away or spiral inwards instead of maintaining a circular trajectory.