DP Physics (first assessment 2025)

Question SPM.2.SL.TZ0.3

Date	May Specimen-2023	Marks available	[Maximum mark: 4]	Reference code	SPM.2.SL.TZ0.3
Level	SL	Paper	2	Time zone	TZ0
Command term	Determine, Draw, Show that	Question number	3	Adapted from	N/A

[Maximum mark: 4]

SPM.2.SL.TZ0.3

A satellite moves around Earth in a circular orbit.

(a)

Draw an arrow on the diagram to represent the direction of the acceleration of the satellite.

[1]

Markscheme

arrow normal to the orbit towards the Earth ✓

(b)

The following data are given:

Mass of Earth, M = 5.97 × 10²⁴ kg
Radius of Earth, R = 6.37 × 10⁶ m
Orbital period of the satellite, T = 5.62 × 10³ s

(b.i)

Kepler’s Third Law of orbital motion states that $T^{2} = k r^{3}$ where $k$ is a constant and $r$ is the orbital radius of the satellite.

Show that $k = \frac{4 π^{2}}{G M}$ .

[1]

Markscheme

use of $v_{orbital} = \frac{2 π r}{T}$ AND either $v_{orbital} = \sqrt{\frac{G M}{r}}$ or $\frac{m v_{orbital}^{2}}{r} = \frac{G M m}{r^{2}}$ correctly manipulated ✓

«to yield $T^{2} = (\frac{4 π^{2}}{G M}) r^{3}$ »

Allow use of ω.

(b.ii)

Determine the height of the satellite above the Earth’s surface.

[2]

Markscheme

$r = \sqrt[3]{\frac{G M T^{2}}{4 π^{2}}} = \sqrt[3]{\frac{6.67 \times 10^{- 11} \times 5.97 \times 10^{24} \times {(5620)}^{2}}{4 π^{2}}}$ ✓

$= 6.83 \times 10^{6}$ «m»

height = « $6.83 \times 10^{6} - 6.37 \times 10^{6} =$ » $4.6 \times 10^{5}$ «m» ✓

Syllabus sections

A. Space, time and motion

D. Fields

D. Fields » D.1 Gravitational fields

A. Space, time and motion » A.2 Forces and momentum

Question SPM.2.SL.TZ0.3

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