IB Physics SL D.1.1 Kepler’s Three Laws of Orbital Motion Question BankPractise IB Physics SL D.1.1 by applying Kepler’s laws to orbital radius, period, planetary motion and changing orbital speed.SyllabusFirst assessment 2025CoursePhysics SLLevelSL
D.1.1—Kepler’s three laws of orbital motion question 1[Maximum number: 3]A satellite powered by solar cells directed towards the Sun is in a polar orbit about the Earth.The satellite is orbiting the Earth at a distance of 6600 km from the centre of the Earth.Determine the orbital period for the satellite.Mass of Earth =6.0×1024 kg=6.0 \times 10^{24} \mathrm{~kg}=6.0×1024 kgShow Answermv2r=GMmr2 leading to T2=4π2r3GMT=5320 «s» \begin{aligned} & \frac{m v^{2}}{r}=G \frac{M m}{r^{2}} & \text { leading to } T^{2}=\frac{4 \pi^{2} r^{3}}{G M} & T=5320 \text { «s» } \end{aligned}rmv2=Gr2Mm leading to T2=GM4π2r3T=5320 «s» Alternative 2 « V=GmEr » =6.67×10−11×6.0×10246600×103 OR 7800 « ms−1 » distance =2πr=2π×6600×103 « m » or 4.15×107 « m » « T=dv=4.15×1077800 » =5300 « S » \begin{aligned} & \text { « } V=\sqrt{\frac{G m_{E}}{r}} \text { » }=\sqrt{\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{6600 \times 10^{3}}} \text { OR } 7800 \text { « } \mathrm{ms}^{-1} \text { » } & \text { distance }=2 \pi r=2 \pi \times 6600 \times 10^{3} \text { « } \mathrm{m} \text { » or } 4.15 \times 10^{7} \text { « } \mathrm{m} \text { » } & \text { « } T=\frac{d}{v}=\frac{4.15 \times 10^{7}}{7800} \text { » }=5300 \text { « } \mathrm{S} \text { » } \end{aligned} « V=rGmE » =6600×1036.67×10−11×6.0×1024 OR 7800 « ms−1 » distance =2πr=2π×6600×103 « m » or 4.15×107 « m » « T=vd=78004.15×107 » =5300 « S » Accept use of ω\omegaω nistead of vAdd to Test
