CAIE A-Level Physics 13 Gravitational Fields

Practise calculating gravitational field strength, force, potential and energy, deriving g = GM/r² and applying Newtonian orbital or satellite models.

Exam points

represent gravitational fields and calculate field strength from force, mass and point-mass relationships
apply Newtonian gravitation to orbital motion and geostationary satellites
derive and use g = GM/r² and explain field variation near Earth
calculate gravitational potential and potential energy and interpret work and sign

Covered syllabus points

13.1 Gravitational field 13.2 Gravitational force between point masses 13.3 Gravitational field of a point mass 13.4 Gravitational potential

[Maximum number: 5]

(a)

Define gravitational field.

[ 1 ]

(b)

A spherical planet can be considered as a point mass at its centre.

[ 4 ]

(i)

On Fig. 1.1, draw gravitational field lines outside the planet to represent the gravitational field due to the planet.

Fig. 1.1

[ 2 ]

(ii)

A satellite is in a circular orbit around the planet.

Explain, with reference to your answer in (b)(i), why the path of the satellite is circular.

[ 2 ]

[Maximum number: 8]

(a)

Define gravitational field strength.

[ 1 ]

(b)

The nearest star to the Sun is Proxima Centauri.

This star has a mass of $2.5 \times 10^{29} \mathrm{~kg}$ and is a distance of $4.0 \times 10^{13} \mathrm{~km}$ from the Sun. The Sun has a mass of $2.0 \times 10^{30} \mathrm{~kg}$ .

[ 5 ]

(i)

State why Proxima Centauri may be assumed to be a point mass when viewed from the Sun.

[ 1 ]

radius/diameter/size (of Proxima Centauri) << /is much less than $4.0 \times 10^{13} \mathrm{~km} /$ separation (of Sun and star)
or
(because) it is a uniform sphere

(ii)

Calculate
1. the gravitational field strength due to Proxima Centauri at a distance of $4.0 \times 10^{13} \mathrm{~km}$ ,
field strength = $\mathrm{Nkg}^{-1}$
2. the gravitational force of attraction between the Sun and Proxima Centauri.
force = N

[ 4 ]

1. field strength $=G M / x^{2}$

\begin{aligned} & =\left(6.67 \times 10^{-11} \times 2.5 \times 10^{29}\right) /\left(4.0 \times 10^{13} \times 10^{3}\right)^{2} \\ & =1.0 \times 10^{-14} \mathrm{Nkg}^{-1} \end{aligned}

2. force = field strength × mass

=1.0 \times 10^{-14} \times 2.0 \times 10^{30}

\begin{aligned} \text { force } & =G M m / x^{2} \\ & =\left(6.67 \times 10^{-11} \times 2.5 \times 10^{29} \times 2.0 \times 10^{30}\right) /\left(4.0 \times 10^{13} \times 10^{3}\right)^{2} \\ & =2.0 \times 10^{16} \mathrm{~N} \end{aligned}

(c)

Suggest quantitatively why it may be assumed that the Sun is isolated in space from other stars.

[ 2 ]

force (of $2 \times 10^{16} \mathrm{~N}$ ) would have little effect on (large) mass of Sun ..... B1
would cause an acceleration of Sun of $1.0 \times 10^{-14} \mathrm{~m} \mathrm{~s}^{-2} /$ very small / negligible acceleration ..... B1
or
> many stars all around the Sun
> net effect of forces/fields is zero

[Maximum number: 10]

(a)

State Newton's law of gravitation.

[ 2 ]

(b)

Use Newton's law of gravitation to show that the gravitational field strength g at a distance r away from a point mass M is given by

g=\frac{G M}{r^{2}} .

[ 2 ]

g=F / m

C1

$F=G M m / r^{2}$

and so
$g=\left[G M m / r^{2}\right] / m=G M / r^{2}$

A1

(c)

The Earth has a mass of $5.98 \times 10^{24} \mathrm{~kg}$ and a radius of $6.37 \times 10^{6} \mathrm{~m}$ .

The Moon has a mass of $7.35 \times 10^{22} \mathrm{~kg}$ and a radius of $1.74 \times 10^{6} \mathrm{~m}$ .
The Earth and the Moon can both be considered as point masses at their centres. Their centres are a distance of $3.84 \times 10^{8} \mathrm{~m}$ apart.

[ 6 ]

(i)

Show that the gravitational field strength at the surface of the Moon due to the mass of the Moon is $1.62 \mathrm{~N} \mathrm{~kg}^{-1}$ .

[ 1 ]

$g=\left(6.67 \times 10^{-11} \times 7.35 \times 10^{22}\right) /\left(1.74 \times 10^{6}\right)^{2}=1.62 \mathrm{~N} \mathrm{~kg}^{-1}$

A1

(ii)

Explain why there is a point X on the line between the centres of the Earth and the Moon where the resultant gravitational field strength due to the Earth and the Moon is zero.

[ 2 ]

(iii)

Calculate the distance x of point X from the centre of the Moon.
x= m

[ 3 ]

distance of X from Earth $=\left(3.84 \times 10^{8}-x\right)$

C1

$(G \times) 7.35 \times 10^{22} / x^{2}=(G \times) 5.98 \times 10^{24} /\left(3.84 \times 10^{8}-x\right)^{2}$

C1

$x=3.8 \times 10^{7} \mathrm{~m}$

A1

[Maximum number: 9]

(a)

Define gravitational potential.

[ 2 ]

(b)

The Earth E and the Moon M can both be considered as isolated point masses at their centres. The mass of the Earth is $5.98 \times 10^{24} \mathrm{~kg}$ and the mass of the Moon is $7.35 \times 10^{22} \mathrm{~kg}$ . The Earth and the Moon are separated by a distance of $3.84 \times 10^{8} \mathrm{~m}$ , as shown in Fig. 2.1.

Fig. 2.1 (not to scale)

P is a point, on the line joining the centres of E and M, where the resultant gravitational field strength is zero. Point P is at a distance x from the centre of the Earth.

[ 7 ]

(i)

Explain how it is possible for the gravitational field strength to be zero despite the presence of two large masses nearby.

[ 2 ]

(ii)

Show that x is approximately $3.5 \times 10^{8} \mathrm{~m}$ .

[ 2 ]

$g \propto M / r^{2}$

C1

$\begin{aligned} 5.98 \times 10^{24} / x^{2}=7.35 \times 10^{22} /\left(3.84 \times 10^{8}-x\right)^{2} \text { leading to } x=3.5 \times 10^{8}(\mathrm{~m}) \end{aligned}$

A1

(iii)

Calculate the gravitational potential $\phi$ at point P .

\phi=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . . \mathrm{Jkg}^{-1}[3]

[ 3 ]

$\begin{aligned} \phi(\text { Earth })=(-) 6.67 \times 10^{-11} \times\left(5.98 \times 10^{24} / 3.5 \times 10^{8}\right) \text { and } \phi(\text { Moon })=(-) 6.67 \times 10^{-11} \times\left(7.35 \times 10^{22} / 0.38 \times 10^{8}\right) \end{aligned}$

C1

$\phi=(-) 6.67 \times 10^{-11} \times\left[\left(5.98 \times 10^{24} / 3.5 \times 10^{8}\right)+\left(7.35 \times 10^{22} / 0.38 \times 10^{8}\right)\right]$

C1

$=-1.3 \times 10^{6} \mathrm{~J} \mathrm{~kg}^{-1}$

A1

All question bank results loaded

Covered syllabus points

13. Gravitational fields question 1

Question (a)

Question (b)

Question (i)

Question (ii)

13. Gravitational fields question 2

Question (a)

Question (b)

Question (i)

Question (ii)

Question (c)

13. Gravitational fields question 3

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)

Question (iii)

13. Gravitational fields question 4

Question (a)

Question (b)

Question (i)

Question (ii)

Question (iii)