9.1 Gravitational fields — Practise

Work through these on your own. Measure every distance from the centre of the planet, keep the minus sign on every gravitational potential, and write the unit of every answer. In the questions that ask for a reason, write your answer in full sentences.

The questions and the answers below come from a set that is checked by computer every time the book is built.

Two spheres, each of mass 1000 kg, are 2.0 m apart. Find the gravitational force between them.
Mars has a mass of 6.42e23 kg and a radius of 3.39e6 m. Find the gravitational field strength at its surface.
The gravitational field strength at the Earth's surface is 9.81 N/kg. Find its value at a distance of 3R from the centre, where R is the Earth's radius.
Using the mass and radius of Mars in question 2, find the gravitational potential at its surface.
A satellite orbits the Earth in a circle of radius 7.0e6 m. Find its orbital speed.
Find the orbital period of the satellite in question 5.
Two satellites orbit the same planet. Their orbital radii are in the ratio 4:1. Find the ratio of their periods.
Find the escape speed from the surface of Mars. Use its mass 6.42e23 kg and radius 3.39e6 m.
A 500 kg probe moves from a point where the gravitational potential is -4.0e7 J/kg to a point where it is -2.0e7 J/kg. Find the energy needed.
A satellite moves in a circular orbit of radius r around a planet of mass M. State which equation you would use to find its speed, and why.
The gravitational potential at the surface of the Earth is -6.25e7 J/kg with an uncertainty of 5.0e5 J/kg. At the orbit it is -5.88e7 J/kg with an uncertainty of 5.0e5 J/kg. Find the potential difference and its absolute uncertainty.
The escape speed from Mars is 5026 m/s. Write it to two significant figures.
The spacecraft orbits at a height of 400 km. Write this height in metres.
A 1200 kg satellite is lifted from the Earth's surface, where the potential is -6.25e7 J/kg, to a higher orbit where the potential is -5.88e7 J/kg. Find the energy needed.

Answers and concise solutions

1. The gravitational force

Answer: $F = \dfrac{GMm}{r^{2}} = \dfrac{6.67 \times 10^{-11} \times 1000 \times 1000}{2.0^{2}} = 1.7 \times 10^{-5}\,\text{N}$ .

2. The field strength

Answer: $g = \dfrac{GM}{r^{2}} = \dfrac{6.67 \times 10^{-11} \times 6.42 \times 10^{23}}{(3.39 \times 10^{6})^{2}} = 3.7\,\text{N}\,\text{kg}^{-1}$ .

3. The field strength

Answer: $g \propto \dfrac{1}{r^{2}}$ , so tripling the distance divides the field strength by 9: $g = \dfrac{9.81}{9} = 1.09\,\text{N}\,\text{kg}^{-1}$ .

4. The gravitational potential

Answer: $V = -\dfrac{GM}{R} = -\dfrac{4.28 \times 10^{13}}{3.39 \times 10^{6}} = -1.3 \times 10^{7}\,\text{J}\,\text{kg}^{-1}$ .

5. The orbital speed

Answer: $v = \sqrt{\dfrac{GM}{r}} = \sqrt{\dfrac{3.98 \times 10^{14}}{7.0 \times 10^{6}}} = 7.5 \times 10^{3}\,\text{m}\,\text{s}^{-1}$ .

6. The period

Answer: $T = \dfrac{2\pi r}{v} = \dfrac{2\pi \times 7.0 \times 10^{6}}{7.54 \times 10^{3}} = 5.8 \times 10^{3}\,\text{s}$ , about 97 minutes.

7. The ratio of the periods

Answer: Since $\dfrac{r^{3}}{T^{2}}$ is constant, $T \propto r^{3/2}$ . The ratio is $4^{3/2} = 8$ , so the periods are in the ratio 8:1.

8. The escape speed

Answer: $v_{\text{esc}} = \sqrt{\dfrac{2GM}{R}} = \sqrt{\dfrac{8.56 \times 10^{13}}{3.39 \times 10^{6}}} = 5.0 \times 10^{3}\,\text{m}\,\text{s}^{-1}$ .

9. The energy needed

Answer: $\Delta V = (-2.0 \times 10^{7}) - (-4.0 \times 10^{7}) = 2.0 \times 10^{7}\,\text{J}\,\text{kg}^{-1}$ , so $E = m\Delta V = 500 \times 2.0 \times 10^{7} = 1.0 \times 10^{10}\,\text{J}$ .

10. A satellite moves in a circular orbit of radius r around a planet of mass M. State which equation you would use to find its speed, and why.

The gravitational force on the satellite is $\dfrac{GMm}{r^{2}}$ .
The centripetal force needed for the circular orbit is $\dfrac{mv^{2}}{r}$ .
Set the two forces equal. The mass of the satellite cancels, so the speed does not depend on it.
Rearranging gives $v = \sqrt{\dfrac{GM}{r}}$ .

11. The potential difference and its uncertainty

Answer: The difference is $(-5.88 \times 10^{7}) - (-6.25 \times 10^{7}) = 3.7 \times 10^{6}\,\text{J}\,\text{kg}^{-1}$ . For a difference the absolute uncertainties add: $5.0 \times 10^{5} + 5.0 \times 10^{5} = 1.0 \times 10^{6}\,\text{J}\,\text{kg}^{-1}$ . So the difference is $(3.7 \pm 1.0) \times 10^{6}\,\text{J}\,\text{kg}^{-1}$ .

12. The escape speed to two significant figures

Answer: $5026\,\text{m}\,\text{s}^{-1}$ is $5.0 \times 10^{3}\,\text{m}\,\text{s}^{-1}$ to two significant figures. The trailing zero is shown in scientific notation so that the two figures are clear.

13. The height in metres

Answer: $400\,\text{km} = 400 \times 10^{3}\,\text{m} = 4.0 \times 10^{5}\,\text{m}$ .

14. The energy needed

Answer: $\Delta V = 3.7 \times 10^{6}\,\text{J}\,\text{kg}^{-1}$ , so $E = m\Delta V = 1200 \times 3.7 \times 10^{6} = 4.4 \times 10^{9}\,\text{J}$ . This is the energy to lift the satellite against gravity. The extra energy to give it orbital speed is not included.