# 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.