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:
[tex]T_2 = 1.40 hours[/tex]
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
[tex]\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}[/tex]
so we know that
for moon
[tex]T_1 = 27.4 days[/tex]
[tex]r_1 = 384,400 km[/tex]
[tex]r_2 = 6370 km[/tex]
now from above formula we have
[tex]\frac{27.4^2}{T_2^2} = \frac{384,400^3}{6370^3}[/tex]
[tex]T_2 = 0.058 days[/tex]
[tex]T_2 = 1.40 hours[/tex]
