Explanation:
It is given that,
The time period of artificial satellite in a circular orbit of radius R is T. The relation between the time period and the radius is given by :
![T^2\propto R^3](https://tex.z-dn.net/?f=T%5E2%5Cpropto%20R%5E3)
The radius of the orbit in which time period is 8T is R'. So, the relation is given by :
![(\dfrac{T}{T'})^2=(\dfrac{R}{R'})^3](https://tex.z-dn.net/?f=%28%5Cdfrac%7BT%7D%7BT%27%7D%29%5E2%3D%28%5Cdfrac%7BR%7D%7BR%27%7D%29%5E3)
![\dfrac{1}{64}=(\dfrac{R}{R'})^3](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B64%7D%3D%28%5Cdfrac%7BR%7D%7BR%27%7D%29%5E3)
![R'=4\times R](https://tex.z-dn.net/?f=R%27%3D4%5Ctimes%20R)
So, the radius of the orbit in which time period is 8T is 4R. Hence, this is the required solution.
Answer:
H2O
Explanation:
PLS MARK ME TO THE BRAINLIST
Grams (g) is much lighter than kilograms (kg)
Answer:
litre.50000665432158900643212lo