Imagine a light planet revolving around a very massive star in a circular orbit of radius r with direction T. On what power of r, will the square of time period depends if the gravitational force of attraction between the planet and the star is proportional to `r^(-5//2)` .

Options

A. The car cannot make a turn without skidding
B. If the car turns at a speed less than 40 km/hr, it will slip down
C. If the car turns at the correct speed of 40 km/hr the force by the road on the car is equal to `(mv^(2))/(r )`
D. If the car turns at the correct speed of 40 km/hr the force by the road on the car is greater than mg as well as greater than `(mv^(2))/(r )`

Answer

B, D

Explanation

For exactly 40 km/hr, car will not slip,
`v = sqrt(r g tan theta)` r decreases as v decreases
`N sin theta = (mv^(2))/(r )` , `N cos theta = m g N = sqrt(((mv^(2))/(r ))^(2) + (mg)^(2))`