IB Physics HL D.1 Gravitational Fields Question Bank

Practise IB Physics HL D.1 by analysing gravitational fields, orbital dynamics, potential energy and extended satellite evidence.

[Maximum number: 13]

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

(a)

State Newton's universal law of gravitation.

[ 2 ]

(b)

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.

(c)

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}

[ 8 ]

(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.

(ii)

The satellite moves from an orbit of radius 1200 km above the Earth to one of radius 2500 km . The mass of the satellite is 45 kg .

Calculate the change in the gravitational potential energy of the satellite.

[ 3 ]

clear use of $\Delta V=\frac{\Delta E}{m}$ and $V=-\frac{G m}{r}$ or $\Delta E=G M m\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right)$ ;
one value of potential energy calculated $\left(2.37 \times 10^{9}\right.$ or $\left.2.02 \times 10^{9}\right)$ ;

Marking guidance:

Award [3] for a bald correct answer.

(iii)

Explain whether the gravitational potential energy has increased, decreased or stayed the same when the orbit changes, as in (c)(ii).

[ 2 ]

increased; further from Earth / closer to infinity / smaller negative value;

Part 2 Properties of a gas
(a) ( Q ) energy transferred between two objects (at different temperatures);
( U ) (total) potential energy and (random) kinetic energy of the molecules/particles (of the gas);
(b) (i) use of area within cycle;
each large square has work value of 250 J ;
estimate $(16 \times 250=) 4000 \mathrm{~J}$ ; (allow 3600 - 4100)
(ii) (work is done by the gas because) area under expansion is greater than that under compression/pressure during expansion is greater than during compression;
(iii) clear attempt to compare two P V values; evaluate two P V values correctly eg $75 \times 80=6000$ and $200 \times 30=6000$ ;
(iv) use of P V=n R T or equivalent;
$1350 / 1330 \mathrm{~K}$ ;
(v) both changes are isochoric/isovolumetric/constant volume changes;

B: temperature/internal energy increases, D: temperature/internal energy decreases;
B: thermal energy/heat input (to system), D: thermal energy/heat output (from system);
B: pressure increases, D: pressure decreases;

[Maximum number: 9]

There is a proposal to place a satellite in orbit around planet Mars.

(a)

Outline what is meant by gravitational field strength at a point.

[ 2 ]

(b)

Newton's law of gravitation applies to point masses. Suggest why the law can be applied to a satellite orbiting Mars.

[ 2 ]

(c)

The satellite is to have an orbital time T equal to the length of a day on Mars. It can be shown that

T^{2}=k R^{3}

where R is the orbital radius of the satellite and k is a constant.

[ 5 ]

(i)

Mars has a mass of $6.4 \times 10^{23} \mathrm{~kg}$ . Show that, for Mars, k is about $9 \times 10^{-13} \mathrm{~s}^{2} \mathrm{~m}^{-3}$ .

[ 3 ]

« $\frac{m v^{2}}{r}=\frac{G M m}{r^{2}}$ hence» $v=\sqrt{\frac{G M}{R}}$ . Also $v=\frac{2 \pi R}{T}$

OR
$m \omega^{2} r=\frac{G M m}{r^{2}} \text { hence } \omega^{2}=\frac{G M}{R^{3}}$

uses either of the above to get $T^{2}=\frac{4 \pi^{2}}{G M} R^{3}$

OR uses $k=\frac{4 \pi^{2}}{G M}$ $k=9.2 \times 10^{-13} / 9.3 \times 10^{-13}$

Unit not required

(ii)

The time taken for Mars to revolve on its axis is $8.9 \times 10^{4} \mathrm{~s}$ . Calculate, in $\mathrm{m} \mathrm{s}^{-1}$ , the orbital speed of the satellite.

[ 2 ]

$\begin{aligned} & R^{3}=\frac{T^{2}}{k}=\frac{\left(8.9 \times 10^{4}\right)^{2}}{9.25 \times 10^{-13}} R=2.04 \times 10^{7}<\mathrm{m} \gg & v=<\omega r=\frac{2 \pi \times 2.04 \times 10^{7}}{89000}=>1.4 \times 10^{3}<\mathrm{ms}^{-1}> \end{aligned}$

OR
$v=\ll \sqrt{\frac{G M}{R}}=\sqrt{\frac{6.67 \times 10^{-11} \times 6.4 \times 10^{23}}{2.04 \times 10^{7}}}=>1.4 \times 10^{3}<\mathrm{ms}^{-1} \gg$

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IB Physics HL D.1 Gravitational Fields Question Bank

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Covered syllabus points

D.1 Gravitational fields question 1

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)

Question (iii)

D.1 Gravitational fields question 2

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)