Paper 3 Gravitational Fields Answers

Use $G=6.67\times 10^{-11}\mathrm{N}{\mathrm{m}}^2\mathrm{k}{\mathrm{g}}^{-2}$.

Q1 Gravitational potential and work

1. $\varphi =-GM/r$. At $r_1$, ${\varphi }_1=-4.98\times 10^7\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. At $r_2$, ${\varphi }_2=-2.49\times 10^7\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. [3]
2. $\Delta U=m({\varphi }_2-{\varphi }_1)=950(2.49\times 10^7)=2.37\times 10^{10}\mathrm{J}$. [3]
3. The potential energy increases, so positive external work must be supplied if kinetic energy is unchanged. [2]
4. Moving a mass from infinity to a point near Earth releases energy; equivalently, work must be done to remove it from the field to infinity. Therefore the potential near Earth is lower than zero. [2]

Q2 Geostationary orbit

1. Circular orbit; above the equator; same direction as Earth’s rotation; period equal to Earth’s rotation period. Any three. [3]
2. Only an equatorial orbit can keep the satellite above a fixed latitude and longitude on Earth’s surface as Earth rotates. An inclined orbit would move north and south relative to the equator. [2]
3. Gravity provides centripetal force:

$$\frac{GMm}{r^2}=m{\omega }^{2}r.$$

Cancel $m$ and rearrange: $GM={\omega }^2{r}^3$, so $r^3=GM/{\omega }^2$. [3] 4. $\omega =2\pi /(8.64\times 10^4)=7.27\times 10^{-5}\mathrm{rad}{\mathrm{s}}^{-1}$. Hence

$$r={\left(\frac{6.67\times 10^{-11}(5.97\times 10^{24})}{(7.27\times 10^{-5}{)}^2}\right)}^{1/3}=4.22\times 10^7\mathrm{m}.$$

[3]