Solve the following examples/numerical problems :
If the height of a satellite completing one revolution around the earth in T seconds in h meters , then what would be the height of a satellite taking 2 √2 T seconds for one revolutions?

Explanation

Time period of satellite (T) = `(2pi(R+h))/(v_(c))`
The height attained by first satellite is given to be `h_(1)` .
For first satellite.
`T=(2pi(R+h))/(v_(c))=2pisqrt((R+h_(1))^(3)/(GM))" "…. (i) `
Let the height attained by second satellite be `h_(2)` .
Given that `T_(2)=2sqrt(2)T`
`therefore` For second satellite ,
`T_(2)=2sqrt(2)T=2pisqrt((R+h_(2))^(3)/(GM))` ....(ii)
Dividing equation (ii) by equation (i) , we get ,
`2sqrt(2)=sqrt((R+h_(2))^(3)/(R+h_(1))^(3))`
Squaring both the sides ,
`8=(R+h_(2))^(3)/((R+h_(1))^(3))`
`therefore2(R+h_(1))=R+h_(2)`
`therefore2R+2h_(1)=R+h_(2)`
`thereforeh_(2)=R+2h_(1)`
The second satellite would attain the height of `(R+2h_(1))` metre.
Alternate method using Kepler's law:
According to Kepler's third law,
`T^(2)propr^(3)`
`therefore(T_(2)^(2))/(T_(1)^(2))=(r_(1)^(3))/(r_(1)^(3))` ....(i)
Given that, `T_(1)=Tand T_(2)=2sqrt(2)T`
`"Also" , r_(1)=R+h_(1)and r_(2)=R+h_(2)`
Substituting in equation (i) ,
`((2sqrt(2)T)/(T))^(2)=[((R+h_(2)))/((R+h_(1)))]^(3)`
`therefore" " 8=[((R+h_(2)))/((R+h_(1)))]^(3)`
`therefore" " root3(8)=((R+h_(2)))/((R+h_(1)))=2`
`therefore2(R+h_(1))=(R+h_(2))`
`therefore" " h_(2)=R+2h_(1)`
The second satellite would attain the height of `(R+2h_(1))` netre.