IB Physics HL D.1.1 Kepler’s Three Laws of Orbital Motion Question Bank

Practise IB Physics HL D.1.1 by solving Kepler-law, orbital-period, planetary-mass and two-body motion problems.

[Maximum number: 6]

This question is in two parts. Part 1 is about gravitational force fields. Part 2 is about properties of a gas.

(a)

A satellite of mass m orbits a planet of mass M. Derive the following relationship between the period of the satellite T and the radius of its orbit R (Kepler's third law).

T^{2}=\frac{4 \pi^{2} R^{3}}{G M}

[ 3 ]

$G \frac{M m}{R^{2}}=\frac{m v^{2}}{R}$ so $v^{2}=\frac{G m}{R}$ ;
$v=\frac{2 \pi R}{T} ;$ $\left(v^{2}=\right) \frac{4 \pi^{2} R^{2}}{T^{2}}=\frac{G m}{R} ;$
or
$G \frac{M m}{R^{2}}=m \omega^{2} R ;$ $\omega^{2}=\frac{4 \pi^{2}}{T^{2}} ;$ $\frac{4 \pi^{2}}{T^{2}}=\frac{G M}{R^{3}} ;$

Marking guidance:

Award [3] to a clear response with a missing step.

(b)

A polar orbiting satellite has an orbit which passes above both of the Earth's poles. One polar orbiting satellite used for Earth observation has an orbital period of $6.00 \times 10^{3} \mathrm{~s}$ .

\begin{array}{ll} \text { Mass of Earth } & =5.97 \times 10^{24} \mathrm{~kg} \\ \text { Average radius of Earth } & =6.37 \times 10^{6} \mathrm{~m} \end{array}

[ 3 ]

(i)

Using the relationship in (b), show that the average height above the surface of the Earth for this satellite is about 800 km .

[ 3 ]

$\quad R^{3}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 6000^{2}}{4 \times \pi^{2}}$ ;

\begin{aligned} & R=7.13 \times 10^{6}(\mathrm{~m}) \\ & h=\left(7.13 \times 10^{6}-6.37 \times 10^{6}\right)=760(\mathrm{~km}) \end{aligned}

Award [3] for an answer of 740 with $\pi$ taken as 3.14.

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D.1.1—Kepler’s three laws of orbital motion question 1

Question (a)

Question (b)

Question (i)