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 1 kg weights 5 cm 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 6360 km, and its mass is 5.97 x 1024kg. What is the difference between the gravitational force on 1 kg at the top of your body, and on 1 kg at your head, and 1 kg 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.