Question

Two satellites, A and B, are in different circular orbits about the earth. The orbital speed of satellite A is three times that of satellite B. Find the ratio ( ) of the periods of the satellites.

Answer 1

Answer:

$\frac{Ta}{Tb} =\frac{1}{27}$

Explanation:

formula for the period of a satellite = 2π.$\frac{r^{3/2} }{\sqrt{GMe}}$

period of satellite A (Ta) = 2π.$\frac{Ra^{3/2} }{\sqrt{GMe}}$

period of satellite B (Tb) = 2π.$\frac{Rb^{3/2} }{\sqrt{GMe}}$

ratio of the periods of the satellites  $\frac{Ta}{Tb}$ = $\frac{2π.\frac{Ra^{3/2} }{\sqrt{GMe}}}{2π.\frac{Rb^{3/2} }{\sqrt{GMe}}}$   (take note that π is shown as $π$)

where

• Me = mass of the earth
• G = universal gravitational constant
• π = 3.142
• all the above are constants and cancel each since they are both at the numerator and denominator
• R = distance from the earth

the equation now becomes

$\frac{Ta}{Tb}$ = $\frac{Ra^{3/2}}{Rb^{3/2}}$  

• the velocity of a satellite = $\frac{2πR}{T}$   (take note that π is shown as $π$)

• rearranging the above R = $\frac{VT}{2π}$   (take note that π is shown as $π$)
• now substituting the above into the equation we have

$\frac{Ta}{Tb}$ = $\frac{(\frac{VaTa}{2π})^{3/2}}{(\frac{VbTb}{2π})^{3/2}}$ (take note that π is shown as $π$)

2π will cancel itself from the numerator and denominator and we have

$\frac{Ta}{Tb}$ = $\frac{(VaTa)^{3/2}}{(VbTb)^{3/2}}$

squaring both sides we have

$\frac{(Ta)^{2}}{(Tb)^{2}}$ = $\frac{(VaTa)^{3}}{(VbTb)^{3}}$

now cross multiplying we have

$Va^{3}Ta^{3}Tb^{2}=Vb^{3}Tb^{3}Ta^{2}$

$Va^{3}Ta=Vb^{3}Tb$

$\frac{Ta}{Tb} =\frac{Vb^{3} }{Va^{3} }$

from the question the velocity of satellite A is 3 times that of satellite B hence Va = 3Vb

$\frac{Ta}{Tb} =\frac{Vb^{3} }{(3Vb)^{3} }$

$\frac{Ta}{Tb} =\frac{Vb^{3} }{3^{3}Vb^{3} }$

$\frac{Ta}{Tb} =\frac{1}{3^{3} }$

$\frac{Ta}{Tb} =\frac{1}{27}$