Centripetal acceleration of Earth around Sun?

1. Dec 17, 2004

psycovic23

New question: centripetal force?

I can't seem to get the right answer...

R = 1.50 x 10^11

and I figure that 1 day/rev = 86400 sec/rev

So: $$\frac {2 \pi 1.50 * 10^{11}}{86400}$$ is the velocity, and then plug that into $$\frac {v^2}{r}$$ but I keep on getting 700 something..what am I doing wrong?

Question 2: At what minimum speed must a roller coaster be traveling when upside down at the top of a circle if the passengers are not to fall out? R = 8.6m

What I think you have to do is
$$mg + F_c = m * \frac {v^2}{R}$$

but I don't know what F would be...or am I approaching this the wrong way?

Last edited: Dec 17, 2004
2. Dec 17, 2004

dextercioby

Whoever told you that the Earth circles (on an ellipse :tongue2: ) the Sun in one day (86400s) is a WACKO!!!!!!!!!!!!! :surprised :surprised

Daniel.

PS.I hope u see the error. :tongue2:

3. Dec 17, 2004

psycovic23

Wow. I'm dumb :tongue2:

4. Dec 17, 2004

psycovic23

*Bump for new question*

5. Dec 18, 2004

futb0l

Hmm - Newton's second law(the mass cancels)... so, the centripetal acceleration has to be bigger than the gravitational acceleration.

$$\frac{v^2}{r} > -g$$

6. Dec 18, 2004

Staff: Mentor

So far, so good. What you call $F_c$ is the force exerted by the coaster on the passenger. The minimum speed of the coaster is that which will make $F_c = 0$. When that force goes to zero, it means that the passengers are beginning to lose contact with the seat.

7. Dec 18, 2004

psycovic23

Ah, thank you very much!