IB Physics HL D 1 Gravitational Fields Topic Practice

$\quad R^{3}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 6000^{2}}{4 \times \pi^{2}}$;

Award [3] for an answer of 740 with $\pi$ taken as 3.14.

clear use of $\Delta V=\frac{\Delta E}{m}$ and $V=-\frac{G m}{r}$ or $\Delta E=G M m\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right)$;
one value of potential energy calculated $\left(2.37 \times 10^{9}\right.$ or $\left.2.02 \times 10^{9}\right)$;

Marking guidance:

$3.5\times10^{8}\ \mathrm{J}$

Award [3] for a bald correct answer.

increased;
further from Earth / closer to infinity / smaller negative value;

$\frac{mv^2}{r}=\frac{GMm}{r^2}$, hence $v=\sqrt{\frac{GM}{R}}$. Also $v=\frac{2\pi R}{T}$.

OR

$m\omega^2r=\frac{GMm}{r^2}$, hence $\omega^2=\frac{GM}{R^3}$.

uses either of the above to get $T^2=\frac{4\pi^2}{GM}R^3$

OR uses $k=\frac{4\pi^2}{GM}$, $k=9.2\times10^{-13}\ /\ 9.3\times10^{-13}$

$$\begin{aligned} R^3 &= \frac{T^2}{k}=\frac{(8.9\times10^4)^2}{9.25\times10^{-13}}\\ R &= 2.04\times10^7\ \mathrm{m}\\ v &= \omega R=\frac{2\pi\times2.04\times10^7}{89000}=1.4\times10^3\ \mathrm{m\,s^{-1}} \end{aligned}$$

OR

$$v=\sqrt{\frac{6.67\times10^{-11}\times6.4\times10^{23}}{2.04\times10^7}}=1.4\times10^3\ \mathrm{m\,s^{-1}}$$