Question 18N.2.HL.TZ0.8

force per unit mass ✔

acting on a small/test/point mass «placed at the point in the field» ✔

Mars is spherical/a sphere «and of uniform density so behaves as a point mass» ✔

satellite has a much smaller mass/diameter/size than Mars «so approximates to a point mass» ✔

«$\frac{m{v}^2}{r}=\frac{GMm}{r^2}$ hence» $v=\sqrt{\frac{GM}{R}}$. Also $v=\frac{2\pi R}{T}$

OR

$m{\omega }^{2}r=\frac{GMm}{r^2}$ 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 × 10⁻¹³ / 9.3 × 10⁻¹³

Unit not required

$R^3=\frac{T^2}{k}=\frac{{\left(8.9\times {10}^4\right)}^2}{9.25\times {10}^{-13}}$  R = 2.04 × 10⁷ «m» ✔

v = «$\omega r=\frac{2\pi \times 2.04\times {10}^7}{89000}=$» 1.4 × 10³ «m s^–1»

OR

v = «$\sqrt{\frac{GM}{R}}=\sqrt{\frac{6.67\times {10}^{-11}\times 6.4\times {10}^{23}}{2.04\times {10}^7}}=$» 1.4 × 10³ «m s^–1» ✔

use of $I\propto \frac{1}{r^2}$ «1.36 × 10³ × $\frac{1}{{1.5}^2}$» ✔

604 «W m^–2» ✔

use of $\frac{600}{4}$ for mean intensity ✔

temperature/K = «$\sqrt[4]{\frac{600}{4\times 5.67\times {10}^{-8}}}=$» 230 ✔

reference to greenhouse gas/effect ✔

recognize the link between molecular density/concentration and pressure ✔

low pressure means too few molecules to produce a significant heating effect

OR

low pressure means too little radiation re-radiated back to Mars ✔

The greenhouse effect can be described, it doesn’t have to be named