Paper 2 Gravitational Fields Answers

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

Q1 Field and potential near a planet

1. $g=GM/R^2=6.67\times 10^{-11}(5.40\times 10^{24})/(6.20\times 10^6{)}^2=9.37\mathrm{N}\mathrm{k}{\mathrm{g}}^{-1}$. [2]
2. $\varphi =-GM/R=-6.67\times 10^{-11}(5.40\times 10^{24})/(6.20\times 10^6)=-5.81\times 10^7\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. [2]
3. Field strength has direction: it is the force direction per unit mass. Potential is energy per unit mass, and energy is scalar, so potentials add algebraically without direction. [2]

Q2 Superposition

1. The two masses produce equal-magnitude fields at P in opposite directions. Field strength is a vector, so the resultant is zero. [2]
2. Each mass contributes a negative scalar potential. Potentials add algebraically, so the total is the sum of two negative values, not zero. [2]
3. The field would point towards the larger mass because its contribution has greater magnitude. The potential would still be the scalar sum of both negative contributions. [2]

Q3 Circular orbit

1. Gravity provides centripetal force:

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

So $v=\sqrt{GM/r}=\sqrt{6.67\times 10^{-11}(5.97\times 10^{24})/(7.20\times 10^6)}=7.44\times 10^3\mathrm{m}{\mathrm{s}}^{-1}$. [3] 2. $T=2\pi r/v=2\pi (7.20\times 10^6)/(7.44\times 10^3)=6.08\times 10^3\mathrm{s}$. [2] 3. Velocity direction changes continuously. Acceleration is towards the centre, provided by gravitational force. [2]