Question 17N.2.SL.TZ0.5

$\frac{m{v}^2}{r}=G\frac{Mm}{r^2}$

leading to T² = $\frac{4{\pi }^2{r}^3}{GM}$

T = 5320 «s»

Alternative 2

«$v=\sqrt{\frac{G{M}_E}{r}}$» = $\sqrt{\frac{6.67\times {10}^{-11}\times 6.0\times {10}^{24}}{6600\times {10}^3}}$ or 7800 «ms^–1»

distance = 2$\pi$r = 2$\pi$ x 6600 x 10³ «m» or 4.15 x 10⁷ «m»

«T = $\frac{d}{v}=\frac{4.15\times {10}^7}{7800}$» = 5300 «s»

Accept use of ω instead of v

T = «$\frac{2.90\times {10}^{-3}}{{\lambda }_{\text{max}}}=$» $\frac{2.90\times {10}^{-3}}{10.1\times {10}^{-6}}$

= 287 «K» or 14 «°C»

Award [0] for any use of wavelength from Sun

Do not accept 287 °C

wavelength of radiation from the Sun is shorter than that emitted from Earth «and is not absorbed by the atmosphere»

infrared radiation emitted from Earth is absorbed by greenhouse gases in the atmosphere

this radiation is re-emitted in all directions «including back to Earth»

peer review

international collaboration

full details of experiments published so that experiments can repeated

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