DP Physics (first assessment 2025)

Question 22N.2.HL.TZ0.b.ii

Date	November 2022	Marks available	[Maximum mark: 2]	Reference code	22N.2.HL.TZ0.b.ii
Level	HL	Paper	2	Time zone	TZ0
Command term	Determine	Question number	b.ii	Adapted from	N/A

b.ii.

[Maximum mark: 2]

22N.2.HL.TZ0.b.ii

Determine the minimum energy required to launch the satellite. Ignore the original kinetic energy of the satellite due to Earth’s rotation.

[2]

Markscheme

change in PE $= 6.67 \times 10^{- 11} \times 6.0 \times 10^{24} \times 8.0 \times 10^{2} (\frac{1}{6.4 \times 10^{6}} - \frac{1}{6.9 \times 10^{6}}) =$ « $3.6 \times 10^{9}$ J» ✓

energy needed = KE + ΔPE = $2.7 \times 10^{10}$ «J» ✓

Allow ECF from 8(b)(i).

Examiners report

Generally, this question was not well done. Most carried out a calculation based on the formula for escape velocity. An opportunity to remind candidates of reading back the stem for the sub-question when answering a second or any subsequent part of it.

Syllabus sections

D. Fields

D. Fields » D.1 Gravitational fields

Question 22N.2.HL.TZ0.b.ii

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