if the velocity of a car is halved, the centripetal force required to keep it in path of constant radius is (multiplied,divided) by what

Question

matt101 · Answer

The equation for centripetal force is:
$$F_{1}=mv^2/r$$
The new velocity is half the old one, i.e. 0.5v:
$$F_{2}=m(0.5v)^2/r$$$$F_{2}=m(0.25v^2)/r$$
You can divide one equation by the other to see the relationship between the two forces. I generally find this easier to do, but you could also isolate for one of the variables that does not change (v, m, or r) and solve...it takes longer, but gives you the same answer.
$$F_{1}=mv^2/r$$---------------$$F_{2}=m(0.25v^2)/r$$
$$F _{1}/F _{2}=1/0.25$$
$$F _{2}=0.25F _{1}$$
The centripetal force at half the velocity is therefore a quarter of what is needed at the original velocity to maintain the same radius.

Hope that helps! You can use a similar procedure to solve your satellite question, you just need to figure out which equations to use.