IB Physics SL D.1 Gravitational fields questions

Exam-style multiple choice and written questions with full markschemes. Check multiple choice answers straight away, or type, photograph or draw a written answer and get it marked.

Type

Multiple choice Written

Difficulty

Easy Medium Hard

Question 1

Multiple choiceHard1 mark

A satellite moves in an elliptical orbit around a planet. Its closest and farthest distances from the planet are $1.0\times10^7\text{ m}$ and $7.0\times10^7\text{ m}$ A satellite in a circular orbit around the same planet has radius $2.0\times10^7\text{ m}$ and period $4.0\text{ d}$ Which pair gives the elliptical orbit’s period, followed by the ratio of its speed at closest approach to its speed at farthest distance?

Show answer and explanation

Answer: C. For the same planet, Kepler’s third law gives $T^2/a^3=\text{constant}$ and Kepler’s second law gives equal swept areas in equal times.

The semi-major axis is $a=(1.0+7.0)\times10^7/2=4.0\times10^7\text{ m}$ Thus $T=4.0(4.0/2.0)^{3/2}=11.3\text{ d}$ which rounds to $11\text{ d}$
At closest and farthest approach, velocity is perpendicular to the radius, so equal area rates give $r_{\rm close}v_{\rm close}=r_{\rm far}v_{\rm far}$ Hence $v_{\rm close}/v_{\rm far}=7$ A: omits the cube in Kepler’s third-law relation, using $T/T_0=\sqrt{a/a_0}$ B: reverses the speed ratio. D: uses $T/T_0=(a/a_0)^3$ omitting the square root. Remember: orbital speed is greater nearer the central body.

The best questions are still waiting

Unlock every question in this topic, then type, photograph or draw your answer and have it marked against the markscheme. Find out where you are dropping marks.

Unlock all questions Compare Student and Student Plus