Aroller coaster car crosses the top of a circular loop-the-loop at twice the critical speed.  what is the ratio of the normal force to the gravitational force?

R = radius of the loop

m = mass of the roller coaster car

for looping the loop critical speed at the top is given as

v₀ = critical speed at the top = sqrt(2gR)

given that :

v = speed of crossing at the top = 2 v₀ = 2  sqrt(2gR)

F = normal force at the top in downward direction

W = weight in downward direction = mg

force equation at the top is given as

F + W = m v²/R

F + mg = m (2  sqrt(2gR))²/R

F + mg = m (4 x 2 gR)/R

F + mg = m (8gR)/R

F + mg = 8 mg

F = 7 mg

F/(mg) = 7

so the ratio comes out to be 7:1

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your answer is, orbit around the nucleus