Question

A satellite in a circular orbit of radius r has period T. A satellite in a nearby orbit with radius r + Δr, where Δr << r , has the very slightly different period T+ ΔT.

a) Show that $\frac{\Delta T}{T}=\frac{3}{2}\frac{\Delta r}{r}$

b) Two earth satellites are in parallel orbits with radii 6700 km and 6701 km. One day they pass each other, 1 km apart, along a line radially outward from the earth. How long will it be until they are again 1 km apart?

Short answer

a) The given expression is proved.

b) They will meet again after 281.05 days

Step-by-step solution

Part(a) Step1: Given information

Two satellites are crossing each other. r and T are the radius of orbit and time period of first satellite whereas r + Δr, and T + ΔT, are the radius of orbit and time period of the second satellite respectively.

Part(a)Step2: Explanation

From Kepler's law

$T=\sqrt{\frac{4{\pi }^2}{GM}}{r}^{3/2}$

Now lets assume $\mathrm{a}=\sqrt{\frac{4{\pi }^2}{GM}}$, the first satellite obeys the $T=a{r}^{3/2}$.
And , for the second satellite as it varies with Δr and ΔT so it will be

$T+\Delta T=a{r}^{3/2}{\left(1+\frac{\Delta r}{r}\right)}^{3/2}$

as $\frac{\Delta r}{r}<<1$

so,

$T+\Delta T=a{r}^{3/2}\left(1+\frac{3}{2}\frac{\Delta r}{r}\right)$

Now subtracting the equation $T=a{r}^{3/2}$for the first satellite:

$\Delta T=a{r}^{3/2}\left(\frac{3}{2}\frac{\Delta r}{r}\right)$

On dividing with the equation for T, we get

$\frac{\Delta T}{T}=\frac{3}{2}\times \frac{\Delta r}{r}$

And proved.

Part(b) Step1: Given information

Two satellites are in parallel orbit.
Radius of orbit of first satellite,r=6700 km
Radius of orbit of second satellite, r + Δr=6701 km
Mass of Earth =5.98 x 10²⁴ kg

Part(b)Step2: Explanation

In part(a) of the problem we have established

$\frac{\Delta T}{T}=\frac{3}{2}\times \frac{\Delta r}{r}$

Substitute values

$$\begin{gathered} \frac{\Delta T}{T}=\frac{3}{2}\left(\frac{1\mathrm{km}}{6700\mathrm{km}}\right) \\ \frac{\Delta T}{T}=2.24\times {10}^{-4} \end{gathered}$$

So after $\frac{1}{2.24\times {10}^{-4}}=4467$periods, they will meet again.

Find the value of One period

$$\begin{gathered} T=\sqrt{\frac{4{\pi }^2}{G\left(5.98\times {10}^{24}\mathrm{kg}\right)}} \\ =5453.3\mathrm{s} \\ =1.51\mathrm{hr} \end{gathered}$$

So they will meet again after 4467 periods = 4467 x 1.51 hours =6745.2 hours

.6745.2 hours =281.05 days .