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Derive an expression to show that for satellites in a circular orbit $r^2 \propto T^2$ where $T$ is the period of orbit and $r$ is the radius of the orbit. - AQA - A-Level Physics - Question 7 - 2017 - Paper 2

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Derive-an-expression-to-show-that-for-satellites-in-a-circular-orbit--$r^2-\propto-T^2$-where-$T$-is-the-period-of-orbit-and-$r$-is-the-radius-of-the-orbit.-AQA-A-Level Physics-Question 7-2017-Paper 2.png

Derive an expression to show that for satellites in a circular orbit $r^2 \propto T^2$ where $T$ is the period of orbit and $r$ is the radius of the orbit.

Worked Solution & Example Answer:Derive an expression to show that for satellites in a circular orbit $r^2 \propto T^2$ where $T$ is the period of orbit and $r$ is the radius of the orbit. - AQA - A-Level Physics - Question 7 - 2017 - Paper 2

Step 1

Derive From Centripetal Force Equation

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Answer

The centripetal force acting on a satellite in a circular orbit is given by:

Fc=mv2rF_c = \frac{m v^2}{r}

where mm is the mass of the satellite, vv is its orbital speed, and rr is the radius of the orbit. This force must equal the gravitational force acting on the satellite:

Fg=GMmr2F_g = \frac{G M m}{r^2}

where GG is the gravitational constant, and MM is the mass of the planet around which the satellite orbits.

Step 2

Combine Equations

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Answer

Setting the centripetal force equal to the gravitational force gives:

mv2r=GMmr2\frac{m v^2}{r} = \frac{G M m}{r^2}

Cancelling mm and rearranging, we find:

v2=GMrv^2 = \frac{G M}{r}

The orbital period TT is related to the orbital speed by the circumference of the orbit:

v=2πrTv = \frac{2 \pi r}{T}

Substituting for vv gives:

(2πrT)2=GMr\left(\frac{2 \pi r}{T}\right)^2 = \frac{G M}{r}

Rearranging leads to:

T2=4π2r3GMT^2 = \frac{4 \pi^2 r^3}{G M}

This shows that r2T2r^2 \propto T^2.

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