CAIE A-Level Physics 13.4.2 Using φ = −GM/r

Practise calculating and rearranging gravitational potential phi = -GM/r for point or spherical masses, comparing distance, mass, sign and units.

Exam points

calculate gravitational potential using ϕ = −GM/r for a point mass
rearrange ϕ = −GM/r to determine mass, distance or potential and maintain the negative sign and units
compare potential at different distances or around different point masses

[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]

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

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

13.4.2—Φ = –GM / r for the gravitational potential in the field due to a point mass question 1