DP Physics (last assessment 2024)

Question 23M.2.HL.TZ2.6

Date	May 2023	Marks available	[Maximum mark: 11]	Reference code	23M.2.HL.TZ2.6
Level	HL	Paper	2	Time zone	TZ2
Command term	Determine, Draw, Estimate, Explain, Outline, Show that, State	Question number	6	Adapted from	N/A

[Maximum mark: 11]

23M.2.HL.TZ2.6

A moon M orbits a planet P. The gravitational field strength at the surface of P due to P is g_P.
The gravitational field strength at the surface of M due to M is g_M.

For M and P: $\frac{radius of M}{radius of P}$ = 0.27 and $\frac{mass of M}{mass of P}$ = 0.055

(a)

Determine $\frac{g_{M}}{g_{P}}$ .

[2]

Markscheme

Work using g ∝ $\frac{m}{r^{2}}$ ✓

$\frac{g_{M}}{g_{P}} = \frac{m_{M}}{m_{P}} {(\frac{r_{P}}{r_{M}})}^{2}$ = 0.75 ✓

(b)

Point O lies on the line joining the centre of M to the centre of P.

The graph shows the variation of gravitational potential V with distance $x$ from the surface of P to O.

The gradient of the graph is zero at point O.

(b.i)

State and explain the magnitude of the resultant gravitational field strength at O.

[2]

Markscheme

g = 0 ✓

As g «= $- \frac{Δ V_{g}}{Δ r}$ which» is the gradient of the graph

OR
As the force of attraction/field strength of P and M are equal ✓

(b.ii)

Outline why the graph between P and O is negative.

[2]

Markscheme

The gravitational field is attractive so that energy is required «to move away from P» ✓

the gravitational potential is defined as 0 at $\infty$ , (the potential must be negative) ✓

(b.iii)

Show that the gravitational potential V_P at the surface of P due to the mass of P is given by V_P = −g_PR_P where R_P is the radius of the planet.

[2]

Markscheme

V_P = $- \frac{G M}{R_{P}}$ AND g_P = $\frac{G M}{{R_{P}}^{2}}$ (at surface) ✓

Suitable working and cancellation of G and M seen ✓

V_P = −g_PR_P

Must see negative sign

(b.iv)

The gravitational potential due to the mass of M at the surface of P can be assumed to be negligible.

Estimate, using the graph, the gravitational potential at the surface of M due to the mass of M.

[2]

Markscheme

« $\frac{V_{M}}{V_{P}}$ = $\frac{g_{M} R_{M}}{g_{P} R_{P}}$ = 0.75 × 0.27» = 0.20 ✓

V_M = «−6.4 × 10⁷ × 0.2 =» «−»1.3 × 10⁷ «J kg⁻¹»✓

(b.v)

Draw on the axes the variation of gravitational potential between O and M.

[1]

Markscheme

Line always negative, of suitable shape and end point below −8 and above −20 unless awarding ECF from b(iv) ✓

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Additional higher level (AHL)

Additional higher level (AHL) » Topic 10: Fields » 10.1 – Describing fields

Additional higher level (AHL) » Topic 10: Fields

Question 23M.2.HL.TZ2.6

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