What Are the Dynamics in a Circular Orbit?

[oldunion]

In the Bohr model of the hydrogen atom, an electron (mass=9.1x10^-31kg) orbits a proton at a distance of 5.3x10^-11m. The proton pulls on the electron with an electric force of 9.2x10^-8N.

I did F=mv^2/r to solve for velocity. units don't match but i got 2.315. Centripetal acceleration next? I know I've made a mistake thus far.


Just came across this question:
What is the acceleration due to gravity of the sun at the distance of the Earth's orbit?

Thats all the info they give you, they want the answer in m/s^2 :uhh:

 

[xelda]

I think you calculated velocity correctly but you left off the scientific notation. Keep the velocity squared and use that to solve for the acceleration.

a = v^2 / r

This will give you the answer in m/s^2

 

[dextercioby]

As for the second part,what formula would u have to use...?

Daniel.

 

[ramollari]

Just came across this question:
What is the acceleration due to gravity of the sun at the distance of the Earth's orbit?

Thats all the info they give you, they want the answer in m/s^2 :uhh:

Hey, by the way don't use again the formula $$g = \frac{v^2}{r_{sun}}$$ here, because you don't know the velocity! If $$R_{Earth}$$ is the radius of the Earth, and we know $$g$$, then express your answer in terms of $$g$$ and the ratio $$R_{Earth}/r_{sun}$$.

 

[dextercioby]

Yes,he does know the velocity.He can approximate the ellipse Earth is making around the sun and determine the velocity to approx 30Km/s...(Besides,even if you didn't know the physics to compute it,it's very well known,because it's tabulated )...

Daniel.

 

[ramollari]

Yes,he does know the velocity.He can approximate the ellipse Earth is making around the sun and determine the velocity to approx 30Km/s...(Besides,even if you didn't know the physics to compute it,it's very well known,because it's tabulated )...

Daniel.

I don't think that is the purpose.
Wait a minute, why does the Earth velocity come up here? It won't do!

 

[dextercioby]

You probably forgot the question.There it is:"What is the acceleration due to gravity of the sun at the distance of the Earth's orbit?"

The acceleration is simply $\frac{v^{2}}{R}$,where v is the orbital velocity of the Earth (namely ~30Km/s) and R is the mean distance Earth-Sun (~150*10^{9}m}.

Daniel.

 

[ramollari]

It is the Earth moving, not the sun. So the formula you gave gives the centripetal acceleration of the Earth moving around the sun.
I understood the question like this: find the value of $$g$$ at the distance the sun is.

 

[dextercioby]

Nope,it's the other way around.It should be the Earth graviting the sun."due to gravity of the sun" ring a bell...?"at the distance of the Earth's orbit"...Does this one ring a bell,then...?

Daniel.

 

[oldunion]

i got the second part. but the first part is incorrect.

a got a v^2 of 5.358x10^12 then i divided this by the radius of 5.3x10^-11= 1.011x10^23 which is not the answer.


edit: nevermind I am being retarded, they asked for it in revs/ second. is this a new equation or just dimensional analysis?

 

[dextercioby]

Revolutions per second,well that's Hertz and normally,in SI revolutions doesn't have a unit...

Daniel.

 

[oldunion]

so how would i go from meters/s^2 to revs/second

 

[dextercioby]

Hold it,one unit id for angular velocity (Hz) and the other is for linear acceleration.There's no way to go from one to the other...There are different quantities...

Daniel.

 

[ramollari]

so how would i go from meters/s^2 to revs/second

$$a = \omega^2r$$

$$\omega = \sqrt{\frac{a}{r}}$$, where $$\omega$$ is in rad/s.

to go to rev/s, you are looking for the frequency:

$$f = \frac{\omega}{2\pi}$$.

 

[oldunion]

no those units don't jive. you would end up with some fraction of #/T where t is the period. tried anyway, 5.99 or 9.42 incorrect

 

[dextercioby]

Okay:What is velocity in m/s and what is the acceleration in m/s^{2}...?

Daniel.

 

[oldunion]

v^2= 5.358x10^12
a=1.011x10^23

 

[dextercioby]

The first # is correct.The second looks good as well...You're done...

Daniel.

 

[oldunion]

The first # is correct.The second looks good as well...You're done...

Daniel.

yes but this doesn't solve my dilemma of how to find the rev/s


5.99x10^21 rad/s (assuming it is in rad/s for whatever reason) x 1 rev/2pirad=9.42x10^21

 

[dextercioby]

Yes,if that $5.99\times 10^{21} rad \ s^{-1}$ is the correct "omega",then te final answer is correct.

Daniel.

 

[oldunion]

something isn't right. if you do the velocity you will get Newtons=kg/m, this works out to m/s^2. The acceleration is therefore inverse seconds...

 

[dextercioby]

No.Trust me,everything is okay with the units.

Daniel.

 

[oldunion]

The computer doesn't like my answer, ill keep at it. But i just worked it out very clearly and when you do acceleration you get #/s. i have another question though.

As a roller coaster car crosses the top of a 30.0m-diameter loop-the-loop, its apparent weight is the same as its true weight. what is cars top speed?

sum of forces=m(v)^2/r obviously the radius is 15m, no mass is apparent and we must find velocity. I know there is a significant in them saying the apparent and actual weights are the same, sum of forces must be zero

 

[dextercioby]

For the first problem,the units are m/s^{2} for LINEAR ACCELERATION and the same for ANGULAR ACCELERATION.It cannot be Hz...

For the second problem,you're on the right track,if i can say so...

Daniel.

 

[oldunion]

the problem asks for rev/s as the final answer though.

the second problem i am stuck. there must be another value i am not seeing because i need one more to solve.

 

[dextercioby]

Yes,but that's not for acceleration...For the 2-nd:What other value do you need...?

Daniel.

 

[oldunion]

Yes,but that's not for acceleration...For the 2-nd:What other value do you need...?

Daniel.

1st problem, significantly lost.

second problem, answer is zero without further values. but zero is not a logical answer