9.1 Gravitational fields — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check. Measure every distance from the centre of the planet, and keep the minus sign when you write a gravitational potential.

Warm-up: remember

A satellite has a mass of 500 kg and a weight of 4500 N. Find the gravitational field strength at its position.
A car goes round a bend of radius 50 m at a speed of $20\,\text{m}\,\text{s}^{-1}$ . Find its centripetal acceleration.
How much energy is needed to lift a 5.0 kg mass through a height of 2.0 m? Use $g = 9.81\,\text{N}\,\text{kg}^{-1}$ .

Answers

$g = \dfrac{W}{m} = \dfrac{4500}{500} = 9.0\,\text{N}\,\text{kg}^{-1}$ .
$a = \dfrac{v^{2}}{r} = \dfrac{20^{2}}{50} = 8.0\,\text{m}\,\text{s}^{-2}$ .
$E = mgh = 5.0 \times 9.81 \times 2.0 = 98\,\text{J}$ .

If question 1 goes wrong, revisit weight and the field strength in Section 3.2. If question 2 goes wrong, revisit circular motion in Section 3.5. If question 3 goes wrong, revisit gravitational potential energy in Section 3.3.

Practise with help

1. Two spheres, each of mass 1000 kg, are 2.0 m apart. Find the gravitational force between them.

Hint

Both masses are the same, so the numerator of Newton's law is $Gm^{2}$ .

Next step

Write $F = \dfrac{GMm}{r^{2}}$ with $M = m = 1000\,\text{kg}$ and $r = 2.0\,\text{m}$ . Square the distance first.

Full working

\begin{aligned} F &= \frac{GMm}{r^{2}} = \frac{6.67 \times 10^{-11} \times 1000 \times 1000}{2.0^{2}} \\ &= \frac{6.67 \times 10^{-5}}{4.0} = 1.7 \times 10^{-5}\,\text{N} \end{aligned}

The force is $1.7 \times 10^{-5}\,\text{N}$ . Gravity is very weak between everyday masses, which is why you do not notice it.

2. Mars has a mass of $6.42 \times 10^{23}\,\text{kg}$ and a radius of $3.39 \times 10^{6}\,\text{m}$ . Find the gravitational field strength at its surface.

Hint

At the surface, the distance $r$ is the radius of the planet.

Next step

Use $g = \dfrac{GM}{r^{2}}$ , and square $3.39 \times 10^{6}$ before you divide.

Full working

\begin{aligned} g &= \frac{GM}{r^{2}} = \frac{6.67 \times 10^{-11} \times 6.42 \times 10^{23}}{(3.39 \times 10^{6})^{2}} \\ &= \frac{4.28 \times 10^{13}}{1.15 \times 10^{13}} = 3.73\,\text{N}\,\text{kg}^{-1} \end{aligned}

The field strength at the surface of Mars is $3.73\,\text{N}\,\text{kg}^{-1}$ . A 70 kg astronaut would weigh 261 N there.

3. The gravitational field strength at the Earth's surface is $9.81\,\text{N}\,\text{kg}^{-1}$ . Find the field strength at a distance of $2R$ from the centre, where $R$ is the Earth's radius.

Hint

You do not need $G$ or the mass of the Earth. The field strength follows an inverse-square law.

Next step

Work out what happens to $g$ when $r$ becomes $2R$ : the distance is doubled, so the field strength is divided by $2^{2}$ .

Full working

g = \frac{GM}{(2R)^{2}} = \frac{GM}{4R^{2}} = \frac{g_{\text{surface}}}{4} = \frac{9.81}{4} = 2.45\,\text{N}\,\text{kg}^{-1}

The field strength at $2R$ is $2.45\,\text{N}\,\text{kg}^{-1}$ , one quarter of the surface value.

4. Find the gravitational potential at the surface of Mars, using the data in question 2.

Hint

The potential is the energy per unit mass, and it is negative.

Next step

Use $V = -\dfrac{GM}{R}$ with the same numbers as question 2, and keep the minus sign.

Full working

\begin{aligned} V &= -\frac{GM}{R} = -\frac{4.28 \times 10^{13}}{3.39 \times 10^{6}} \\ &= -1.26 \times 10^{7}\,\text{J}\,\text{kg}^{-1} \end{aligned}

The potential is $-1.26 \times 10^{7}\,\text{J}\,\text{kg}^{-1}$ . Moving a 1 kg mass from the surface to infinity needs $1.26 \times 10^{7}\,\text{J}$ .

5. A satellite orbits the Earth in a circle of radius $7.0 \times 10^{6}\,\text{m}$ . Find its orbital speed and its period.

Hint

Gravity provides the centripetal force, so set the two force expressions equal.

Next step

Use $v = \sqrt{\dfrac{GM}{r}}$ with $GM = 3.98 \times 10^{14}\,\text{m}^{3}\,\text{s}^{-2}$ , then divide the circumference by the speed.

Full working

\begin{aligned} v &= \sqrt{\frac{GM}{r}} = \sqrt{\frac{3.98 \times 10^{14}}{7.0 \times 10^{6}}} \\ &= \sqrt{5.69 \times 10^{7}} = 7.5 \times 10^{3}\,\text{m}\,\text{s}^{-1} \end{aligned}

T = \frac{2\pi r}{v} = \frac{2\pi \times 7.0 \times 10^{6}}{7.5 \times 10^{3}} = 5.9 \times 10^{3}\,\text{s}

The speed is $7.5\,\text{km}\,\text{s}^{-1}$ and the period is 5900 s, or about 98 minutes.

6. Find the escape speed from the Moon. The mass of the Moon is $7.35 \times 10^{22}\,\text{kg}$ and its radius is $1.74 \times 10^{6}\,\text{m}$ .

Hint

At escape the kinetic energy equals the energy needed to reach infinity.

Next step

Use $v_{\text{esc}} = \sqrt{\dfrac{2GM}{R}}$ . Calculate $2GM$ first, because the factor of 2 is often dropped.

Full working

\begin{aligned} v_{\text{esc}} &= \sqrt{\frac{2GM}{R}} = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 7.35 \times 10^{22}}{1.74 \times 10^{6}}} \\ &= \sqrt{5.63 \times 10^{6}} = 2.4 \times 10^{3}\,\text{m}\,\text{s}^{-1} \end{aligned}

The escape speed from the Moon is $2.4\,\text{km}\,\text{s}^{-1}$ , about one fifth of the escape speed from the Earth.

7. A 200 kg satellite is moved from the Earth's surface, where the potential is $-6.25 \times 10^{7}\,\text{J}\,\text{kg}^{-1}$ , to a higher orbit where the potential is $-5.00 \times 10^{7}\,\text{J}\,\text{kg}^{-1}$ . Find the energy needed.

Hint

The energy depends on the potential difference, not on either potential alone.

Next step

Find $\Delta V$ by subtracting the two negative values carefully. Then multiply by the mass.

Full working

\Delta V = (-5.00 \times 10^{7}) - (-6.25 \times 10^{7}) = 1.25 \times 10^{7}\,\text{J}\,\text{kg}^{-1}

E = m\,\Delta V = 200 \times 1.25 \times 10^{7} = 2.5 \times 10^{9}\,\text{J}

The energy needed is $2.5 \times 10^{9}\,\text{J}$ .

8. Explain why an astronaut in orbit feels weightless even though the gravitational field strength there is $8.69\,\text{N}\,\text{kg}^{-1}$ .

Hint

Weightlessness is about the reading on a balance, not about the force of gravity.

Next step

Say what force acts on the astronaut and the spacecraft, and what happens to both of them because of it.

Full working

Gravity still acts on the astronaut, with a force of about 608 N. The astronaut and the spacecraft are both falling towards the Earth with the same acceleration, because gravity provides the centripetal force for the orbit. The astronaut does not press on the floor of the spacecraft, so a balance would read zero. This is free fall, not a lack of gravity.