How much time would a satellite in an orbit at a height of 35780 km above the earth's surface take to complete one revolution around the earth , if the mass of the earth were four times its original mass ?

Explanation

It is given that the height of the satellie above the earth's surface is 35780 km . Satellites revolving in an orbit at this height above the earth's surface are geosynchronous satellites. i.e , time period of such satellites is around 24 hours. `(T_(1)=24"hours")`
we know that ,
`v_(c)=sqrt((GM)/((R+h)))" " (i)`
`"and "T=(2pi(R+h))/(v_(c))" " (ii)`
From equations (i) and (ii) , we can conclude that,
`T=2pisqrt(((R+h)^(3))/(GM))`
`rArrTprop(1)/(sqrt(M))`
`rArr(T_(1))/(T_(2))=sqrt(M_(2))/sqrt(M_(1))`
Given that , `M_(2)=4M_(1)`
`therefore" " (T_(1))/(T_(2))=sqrt(4M_(1))/sqrt(M_(1))=2`
`therefore" " T_(2)=(T_(1))/(2)=(24)/(2)=12` hours
Hence , the satellite would take approximately 12 hours to compelete one revolution around the earth.