# A-level Physics (Advancing Physics)/Gravitational Forces/Worked Solutions

1. Jupiter orbits the Sun at a radius of around 7.8 x 1011m. The mass of Jupiter is 1.9 x 1027kg, and the mass of the Sun is 2.0 x 1030kg. What is the gravitational force acting on Jupiter? What is the gravitational force acting on the Sun?

$F_{grav} = \frac{-GMm}{r^2} = \frac{-6.67 \times 10^{-11} \times 2 \times 10^{30} \times 1.9 \times 10^{27}}{(7.8 \times 10^{11})^2} = -4.17 \times 10^{23}\mbox{ N}$

2. The force exerted by the Sun on an object at a certain distance is 106N. The object travels half the distance to the Sun. What is the force exerted by the Sun on the object now?

$\frac{1}{\left (\frac{1}{2}\right )^2} = 4$

So, the new force is 4 MN.

3. How much gravitational force do two 1kg weights 5cm apart exert on each other?

$F_{grav} = \frac{-GMm}{r^2} = \frac{-6.67 \times 10^{-11} \times 1 \times 1}{0.05^2} = -2.67 \times 10^{-8}\mbox{ N}$

In other words, ordinary objects exert negligible gravitational force.

4. The radius of the Earth is 6360km, and its mass is 5.97 x 1024kg. What is the difference between the gravitational force on 1kg at the top of your body, and on 1kg at your head, and 1kg at your feet? (Assume that you are 2m tall.)

$\Delta F_{grav} = GMm\left ( \frac{1}{6360000^2} - \frac{1}{6360002^2}\right ) = (1.55 \times 10^{-20})GMm = 1.55 \times 10^{-20} \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 1 = 6.19 \mbox{ }\mu\mbox{N}$

This is why it is acceptable to approximate the acceleration due to gravity as constant over small distances.