Forces of gravity on protons and electrons

In summary: I might be that you are not using the right values for the constants. Check to see if it wants you to use the exact values or if it wants you to round them. Also, make sure your units are consistent. The value you get for r should be in meters, since that is the unit of distance in the equation. In summary, the conversation discusses the problem of finding the distance between an electron and a proton so that its acceleration due to gravity is equivalent to that of a freely falling object on Earth's surface. It also explores the possibility of considering both gravitational and electrical forces in the calculation and suggests using the equations F=ma and F=Gm1m2/r^2. The conversation also mentions the need
  • #1
footballxpaul
30
0
Homework Statement

1.Assuming that only gravity is acting on it, how far does an electron have to be from a proton so that its acceleration is the same as that of a freely falling object at the Earth's surface

2.Suppose the Earth were made only of protons but had the same size and mass it presently has. What would be the acceleration of an electron released at the surface?

3.Is it necessary to consider the gravitational attraction as well as the electrical force?
yes or no?


The Attempt at a Solution



i have read through my chapter and none of the examples help me at all, I thought #1 could be zero but the online homework thing rejected that. I don't know if you could help me with the answer and also give a short explain why that would be greatly appreciated! Thankyou!
 
Physics news on Phys.org
  • #2
footballxpaul said:
Homework Statement

1.Assuming that only gravity is acting on it, how far does an electron have to be from a proton so that its acceleration is the same as that of a freely falling object at the Earth's surface

2.Suppose the Earth were made only of protons but had the same size and mass it presently has. What would be the acceleration of an electron released at the surface?

3.Is it necessary to consider the gravitational attraction as well as the electrical force?
yes or no?


The Attempt at a Solution



i have read through my chapter and none of the examples help me at all, I thought #1 could be zero but the online homework thing rejected that. I don't know if you could help me with the answer and also give a short explain why that would be greatly appreciated! Thankyou!

1) It's asking for when the acceleration = 9.81 m/s^2. You know about F = ma, right? This applies to all forces, including electrical ones. Mass, acceleration, and the two charges are known (proton's charge = electron's charge). Using what two formulas will give you everything but the distance (which can be easily solved for)?

2) First, find the number of protons that can make up Earth. There will be a total charge here. The net force ( F = ma) will be the sum of TWO force vectors in the same direction, the electrical force and the gravitational force. The distance in your equations will be the radius of the earth.

3) Think. Everything with a charge has an electrical field. Everything with mass has a gravitational field.
 
  • #3
You need to get some equations and show us an attempt. For example, the force between two charges is F=k*q1*q2/r^2, F=ma, the acceleration at the Earth's surface is 9.8*m/s^2. Try and put those together to answer the first one.
 
  • #4
for the first one I used f=ma and f=(Gm1m2)/r^2, I am pretty sure the masses could be assumed they are close to zero so the answer would be 1?
 
  • #5
footballxpaul said:
for the first one I used f=ma and f=(Gm1m2)/r^2, I am pretty sure the masses could be assumed they are close to zero so the answer would be 1?

nevermind its wrong let me try again
 
  • #6
footballxpaul said:
for the first one I used f=ma and f=(Gm1m2)/r^2, I am pretty sure the masses could be assumed they are close to zero so the answer would be 1?

You assume the net force (m*a) is only comprised of the gravitational force. What other force could possible contribute to a nonzero net force?

And zero divided by anything is always zero, so 1 would still be wrong.
 
  • #7
F=G*m1*m2/r^2 is the force due to gravity. You want the force due to the electric charge. And you do need to know the masses of the electron and proton. You can't assume them away because they are small. Better look them up.
 
  • #8
Elbobo said:
You assume the net force (m*a) is only comprised of the gravitational force. What other force could possible contribute to a nonzero net force?

And zero divided by anything is always zero, so 1 would still be wrong.

so would I do... Fe +Fg = ma? and solve from there?
 
  • #9
footballxpaul said:
so would I do... Fe +Fg = ma? and solve from there?

You could. But the force due to gravity is completely insignificant compared with the electrical force. You should check this but ignore it for now.
 
  • #10
Dick said:
You could. But the force due to gravity is completely insignificant compared with the electrical force. You should check this but ignore it for now.

ok but with the electrical force they don't list either of the charge values, do I need them or are they the same value, ect?
 
  • #11
footballxpaul said:
ok but with the electrical force they don't list either of the charge values, do I need them or are they the same value, ect?

If they don't list them then you'll need to look them up. They are both the same but with opposite signs.
 
  • #12
With my online system, the website provides the constants that it wants us to use. Check to see if your system already has the constants just in case you use a number too exact or rounded (mine was in the Help section).
 
  • #13
i did...

(k*q1*q2)/(m*9.8) = r^2

still couldn't find the answer but is this the right direction it it f=ma and fe together, big thing also is you can take a negative square root... i am to confused
 
  • #14
What's wrong with that equation? In this case, positive and negative charges mean nothing, so don't make the force negative.
 
  • #15
Ignore the sign. That just tells you which direction the force is acting. It doesn't affect the magnitude of the acceleration. Take the square root of the magnitude.
 
  • #16
footballxpaul said:
i did...

(k*q1*q2)/(m*9.8) = r^2

still couldn't find the answer but is this the right direction it it f=ma and fe together, big thing also is you can take a negative square root... i am to confused

ok still isn't working... one thing for the mass, do i add both the mass of the electron and proton together for m cause that is what I am doing and is the only thing I can think of that I am doing wrong.
 
  • #17
No, only the mass of the electron needs to be calculated.

However, maybe the system is so exact as to want you to factor in the force of gravity? I'm not sure how much it will affect your answer (if at all), so maybe you will have to use Fe + Fg = ma.
 
  • #18
Elbobo said:
No, only the mass of the electron needs to be calculated.

However, maybe the system is so exact as to want you to factor in the force of gravity? I'm not sure how much it will affect your answer (if at all), so maybe you will have to use Fe + Fg = ma.

nope did that this online system(my homework site not this site) is jacked up, Thanks all for helping though
 
  • #19
footballxpaul said:
nope did that this online system(my homework site not this site) is jacked up, Thanks all for helping though

Fg would only affect your answer if you were carrying the results out to about 40 decimal places. Hope your homework site feels better soon.
 
  • #20
hi,1st post:smile: ..i m learning too

izzit by taking the nett forces?
Fg + Fe = ma

i.e

Gm1m2/r^2 + Q1Q2/4(pi)(epsilon zero)r^2 = ma

epsilon zero = permittivity of vacuum
 
  • #21
A|bert said:
hi,1st post:smile: ..i m learning too

izzit by taking the nett forces?
Fg + Fe = ma

i.e

Gm1m2/r^2 + Q1Q2/4(pi)(epsilon zero)r^2 = ma

epsilon zero = permittivity of vacuum

Sure, but the first term is so much smaller than the second, it's ridiculous. Drop it.
 
  • #22
Dick said:
F=G*m1*m2/r^2 is the force due to gravity. You want the force due to the electric charge. And you do need to know the masses of the electron and proton. You can't assume them away because they are small. Better look them up.

That's not what I got out of the original problem statement:

footballxpaul said:
1.Assuming that only gravity is acting on it, how far does an electron have to be from a proton so that its acceleration is the same as that of a freely falling object at the Earth's surface

:wink:
 
  • #23
gabbagabbahey said:
That's not what I got out of the original problem statement:



:wink:

OOOOOOPS. Apparently my reading skills are even worse than my math skills. Thanks for catching that.
 
  • #24
Dick said:
OOOOOOPS. Apparently my reading skills are even worse than my math skills. Thanks for catching that.

No worries m8; I make mistakes like that all the time :smile:
 

1. How does gravity affect protons and electrons?

Gravity affects protons and electrons in the same way as it affects any other objects with mass. The more massive an object is, the stronger its gravitational pull will be. Since protons and electrons have very small masses, their gravitational interaction with each other is usually negligible.

2. Do protons and electrons experience the same amount of gravitational force?

No, protons and electrons do not experience the same amount of gravitational force. This is because protons have a much greater mass than electrons, so they will have a stronger gravitational pull. However, the effects of gravity on protons and electrons are usually very small and are often overshadowed by other forces.

3. How does the distance between protons and electrons affect the force of gravity?

The force of gravity between protons and electrons is directly proportional to the distance between them. This means that as the distance between protons and electrons increases, the force of gravity between them decreases. This is because the gravitational force follows an inverse square law, where the force decreases by a factor of four as the distance doubles.

4. What other factors can influence the forces of gravity on protons and electrons?

Aside from distance, the only other factor that can influence the forces of gravity on protons and electrons is mass. As mentioned earlier, the more massive an object is, the stronger its gravitational pull will be. So, if the masses of protons and electrons were to change, it would also affect the force of gravity between them.

5. Can the forces of gravity on protons and electrons be cancelled out?

Yes, the forces of gravity on protons and electrons can be cancelled out. This can happen if there is an equal and opposite force acting on them, such as the electromagnetic force. In this case, the two forces will cancel each other out, resulting in no net force on the protons and electrons.

Similar threads

Gravity's Pull: How Far is the Proton Above the Electron?
    • Introductory Physics Homework Help
Replies
2
Views
657
Electron bound to proton by gravity
    • Introductory Physics Homework Help
Replies
3
Views
388
Work to move electron/proton between the plates of a charged capacitor?
    • Introductory Physics Homework Help
Replies
22
Views
3K
Solve Columb's Law Problem: Helium Nucleus & Proton Acceleration
    • Introductory Physics Homework Help
Replies
1
Views
714
I Can proton/neutron decay be avoided in some conditions?
    • High Energy, Nuclear, Particle Physics
Replies
2
Views
138
What is the correct velocity of an electron in a bound orbit around a proton?
    • Introductory Physics Homework Help
Replies
2
Views
940
An electron & a proton are each placed in an electric field
    • Introductory Physics Homework Help
Replies
1
Views
2K
Work done by a force down a ramp
    • Introductory Physics Homework Help
Replies
11
Views
1K
Electron and proton penetration
    • Introductory Physics Homework Help
Replies
7
Views
2K
The effect of a stationary proton on a stationary electron
    • Introductory Physics Homework Help
Replies
2
Views
2K

Hot Threads

Recent Insights

Back
Top