Question

The time period of a satellite of Earth is 5 hours. If the separation between earth and satellite is increased to 4 times the previous value calculate the new time period.

Answer 1

According to Kepler’s law of planetary motion T2 ? R3 ? T2 = T1 (R2/R1­)3/2 = 5 x [4R/R]3/2 = 5 x 23 = 40 hour

Answer 2

"As per the Kepler’s third law of Planetary motion the relation between radius(R) and time period (T) is given by

$(\frac { To }{ Tn } )2=(\frac { Ro }{ Rn } )3\Rightarrow (\frac { To }{ Tn } )=(\frac { Ro }{ Rn } )\frac { 3 }{ 2 }$  here To  is old time period and Tn is new time period

$\Rightarrow (\frac { To }{ Tn }) =(\frac { 1 }{ 4 } )\frac { 3 }{ 2 }$

{ Since we know that the distance between the earth and the satellite is quadrupled the new radial distance from earth becomes 4 times the previous radial distance i.e., Rn= 4Ro}

$\Rightarrow\frac { To }{ Tn } =(\frac { 1 }{ 64 } )\frac { 1 }{ 2 }$

$\Rightarrow\frac { To }{ Tn }= \frac { 1 }{ 8 }since Ta is 5hrs (given)$

$\RightarrowTn = 5 \times 8=40 hrs$"