ive been playing with this one for some time now.
my problem is with finding the Radius of a planet if yur given that for every 3600m in tangential, there is 2m in vertical...
ive found a way, but i think my approximations are too course (my final answer is v=5km/s)
any help woudl be greatly appreciated.
(GR9677 #22)
(GR9677 #22) Solution
Hey thermalEquilibrium, I would be happy to help especially since your posts have helped many other members. Thanks.
Yes, I believe the problem is with your approximations for the radius R. Based on your answer, it appears like you are trying to use the small angle approximations, which I imagine can be done but you need to be careful. Anyway, the approximations become very simple if you draw out the right triangle described in the problem and use the Pythagoras theorem:
R^2+3600^2 = (R+2)^2
Which leads to 3600^2 = 4R+4 which shows that R can be very accurately approximated as:
R = 3600^2/4 (meters)
As you probably already know that once you have a good approximation for R then the problem is simple as you just solve the equation where you set the acceleration due to gravity equal to the centripetal acceleration required to keep the golf ball in a circular orbit. Hence:
.4*g=v^2/R
Since we know what g is and have a good approximation for R we can solve for v.
Yes, I believe the problem is with your approximations for the radius R. Based on your answer, it appears like you are trying to use the small angle approximations, which I imagine can be done but you need to be careful. Anyway, the approximations become very simple if you draw out the right triangle described in the problem and use the Pythagoras theorem:
R^2+3600^2 = (R+2)^2
Which leads to 3600^2 = 4R+4 which shows that R can be very accurately approximated as:
R = 3600^2/4 (meters)
As you probably already know that once you have a good approximation for R then the problem is simple as you just solve the equation where you set the acceleration due to gravity equal to the centripetal acceleration required to keep the golf ball in a circular orbit. Hence:
.4*g=v^2/R
Since we know what g is and have a good approximation for R we can solve for v.

 Posts: 7
 Joined: Fri Nov 04, 2005 5:18 pm
I think there's a way around (no pun intended) calculating the radius that worked for me. Maybe I just got lucky, but this might spare you a little time on the test...
You know that the acceleration downward is always 0.4g =~4m/ss. Over the course of such a fall, the average speed should then be 2m/s. So every second, you need to move horizontally fast enough that you have 2 extra meters of space below you.
The curvature is already nicely provided: 3.6 km for every 2m dropoff. Therefore, (c) 3.6 km/s.
I think v^2/r is built into this somewhere, but when I was screaming through this test and timing myself, it seemed a little more intuitive and economical to me.
You know that the acceleration downward is always 0.4g =~4m/ss. Over the course of such a fall, the average speed should then be 2m/s. So every second, you need to move horizontally fast enough that you have 2 extra meters of space below you.
The curvature is already nicely provided: 3.6 km for every 2m dropoff. Therefore, (c) 3.6 km/s.
I think v^2/r is built into this somewhere, but when I was screaming through this test and timing myself, it seemed a little more intuitive and economical to me.