Question

Use Kepler’s laws and the period of the Moon (27.4 d) to determine the period of an artificial satellite orbiting very near the Earth’s surface.

Answer 1

As according to Kepler 's law

T =(4π²r³/ Gm)^1/2

here r= distance  from earth center to satellite = 6400km = 6400000m

G = earth's gravitational constant= 6.67×10^-11

m = mass of earth= 5.98 ×10^24

so T =[ { 4× (3.1416)²×(6400000)³}/ {(6.67×10^-11)×( 5.98 ×10^24)} ]^1/2

T= 5133 sec

Answer 2

Answer:

$T_2 = 1.40 hours$

Explanation:

As per kepler's law of time period we know that

square of time period is proportional to the cube of the distance

so here we can say

$\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}$

so we know that

for moon

$T_1 = 27.4 days$

$r_1 = 384,400 km$

$r_2 = 6370 km$

now from above formula we have

$\frac{27.4^2}{T_2^2} = \frac{384,400^3}{6370^3}$

$T_2 = 0.058 days$

$T_2 = 1.40 hours$