CAIE A-Level Physics A2: Gravitational Potential

[Maximum number: 10]

(a)

Define gravitational potential at a point.

[ 2 ]

(b)

A satellite X , of mass M, orbits a planet at a constant distance 4 R from the centre of the planet, as shown in Fig. 1.1.

Fig. 1.1 (not to scale)

A second satellite Y, of mass 2 M, orbits the planet with orbital radius R.
The gravitational potential at X due to the planet is $-\Phi$ . The planet is a uniform sphere.

[ 8 ]

(i)

Explain why the gravitational potential at X is negative.

[ 2 ]

(ii)

State an expression, in terms of $\Phi$ , for the gravitational potential at Y due to the planet.
gravitational potential =

[ 2 ]

magnitude of potential shown as $4 \Phi$
potential negative and shown as a multiple of $-\Phi$ [potential = $-4 \Phi$ if fully correct]

(iii)

Complete Table 1.1 by giving expressions, in terms of some or all of M, R and $\Phi$ , for the quantities indicated for each of the satellites X and Y .

Table 1.1

[ 4 ]

field strength at $\mathrm{X}: ~ \Phi / 4 R$
field strength at $\mathrm{Y}: \quad 4 \Phi / R$
potential energy at X : $-M \Phi$
potential energy at Y : $-8 M \Phi$

[Maximum number: 5]

(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.

[ 3 ]

(i)

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 ]

$\begin{aligned} \phi(\text { Earth })=(-) 6.67 \times 10^{-11} \times\left(5.98 \times 10^{24} / 3.5 \times 10^{8}\right) \text { and } \phi(\text { Moon })=(-) 6.67 \times 10^{-11} \times\left(7.35 \times 10^{22} / 0.38 \times 10^{8}\right) \end{aligned}$ $\phi=(-) 6.67 \times 10^{-11} \times\left[\left(5.98 \times 10^{24} / 3.5 \times 10^{8}\right)+\left(7.35 \times 10^{22} / 0.38 \times 10^{8}\right)\right]$ $=-1.3 \times 10^{6} \mathrm{~J} \mathrm{~kg}^{-1}$

[Maximum number: 8]

(a)

Define gravitational potential at a point.

[ 2 ]

(b)

The Moon may be considered to be an isolated uniform sphere of mass $7.3 \times 10^{22} \mathrm{~kg}$ and radius $1.7 \times 10^{6} \mathrm{~m}$ .
Calculate the gravitational potential at the surface of the Moon. Give a unit with your answer.
gravitational potential = unit

[ 2 ]

$\begin{aligned} \phi =-G M / r =-\left(6.67 \times 10^{-11} \times 7.3 \times 10^{22}\right) /\left(1.7 \times 10^{6}\right) \end{aligned}$ $=-2.9 \times 10^{6} \mathrm{~J} \mathrm{~kg}^{-1}$

(c)

An isolated uniform spherical planet has gravitational potential $\phi$ at its surface.

A particle of mass m is projected vertically upwards from the surface. The particle is given just enough kinetic energy to travel to an infinite distance away from the planet, escaping from the gravitational pull of the planet, without any additional work being done on it.

[ 3 ]

(i)

Determine an expression, in terms of m and $\phi$ , for the gravitational potential energy $E_{\mathrm{P}}$ of the particle at the surface of the planet.

E_{P}=

[ 1 ]

$E_{\mathrm{P}}=m \phi$

(ii)

Show that the speed v at which the particle is projected upwards from the surface of the planet is given by

v=\sqrt{-2 \phi}

[ 2 ]

$1 / 2 m v^{2}+m \phi=0$
correct algebra leading to $v=\sqrt{ }(-2 \phi)$

(d)

A particle is moving upwards at the surface of the Moon.

Use your answer in (a)(ii) and the expression in (b)(ii) to determine the minimum speed of this particle that will result in it escaping from the gravitational pull of the Moon.
speed = $\mathrm{m} \mathrm{s}^{-1}$

[ 1 ]

$\begin{aligned} \text { speed } =\sqrt{ }\left(2 \times 2.9 \times 10^{6}\right) =2400 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned}$

[Maximum number: 3]

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.

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]

$\begin{aligned} \phi(\text { Earth })=(-) 6.67 \times 10^{-11} \times\left(5.98 \times 10^{24} / 3.5 \times 10^{8}\right) \text { and } \phi(\text { Moon })=(-) 6.67 \times 10^{-11} \times\left(7.35 \times 10^{22} / 0.38 \times 10^{8}\right) \end{aligned}$ $\phi=(-) 6.67 \times 10^{-11} \times\left[\left(5.98 \times 10^{24} / 3.5 \times 10^{8}\right)+\left(7.35 \times 10^{22} / 0.38 \times 10^{8}\right)\right]$ $=-1.3 \times 10^{6} \mathrm{~J} \mathrm{~kg}^{-1}$

CAIE A-Level Physics A2 13.4 Gravitational Potential Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (i)

Question (ii)

Question (iii)

Question 2

Question (a)

Question (b)

Question (i)

Question 3

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)

Question (d)

Question 4