CAIE A-Level Physics 13.2 Gravitational Force Between Point Masses

Practise applying Newton’s gravitational law to point masses and uniform spheres, analysing circular or geostationary orbits and calculating orbital quantities.

Exam points

apply Newton’s law of gravitation to point masses and uniform spheres
analyse circular and geostationary orbits by equating gravitational and centripetal effects
calculate orbital quantities and interpret geostationary conditions

Covered syllabus points

13.2.1—That, for a point outside a uniform sphere, the mass of the sphere may be 13.2.2—Newton’s law of gravitation F = Gm1m2 / r2 for the force between two point 13.2.3—Circular orbits in gravitational fields by relating the gravitational force 13.2.4—A satellite in a geostationary orbit remains at the same point above the

[Maximum number: 1]

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 $2.0 \times 10^{30} \mathrm{~kg}$ .

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

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

[Maximum number: 1]

An isolated uniform conducting sphere has mass M and charge Q.

The gravitational field strength at the surface of the sphere is g.
The electric field strength at the surface of the sphere is E.

Show that the numerical value of $\alpha$ is $1.35 \times 10^{20} \mathrm{~kg}^{2} \mathrm{C}^{-2}$ .

$\begin{aligned} \alpha=1 /\left(4 \pi \times 6.67 \times 10^{-11} \times 8.85 \times 10^{-12}\right)=1.35 \times 10^{20}\left(\mathrm{~kg}^{2} \mathrm{C}^{-2}\right) \text { or } \alpha=\left(8.99 \times 10^{9}\right) /\left(6.67 \times 10^{-11}\right)=1.35 \times 10^{20}\left(\mathrm{~kg}^{2} \mathrm{C}^{-2}\right) \end{aligned}$

A1

[Maximum number: 2]

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

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.

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Covered syllabus points

13.2 Gravitational force between point masses question 1

13.2 Gravitational force between point masses question 2

13.2 Gravitational force between point masses question 3