Any physics genius' out there?

[nmmiles]

A 1.10 -diameter lead sphere has a mass of 5400 kg. A dust particle rests on the surface of the sphere, which itself lies on the surface of the earth.

What is the ratio of the gravitational force of the sphere on the dust particle to the weight of the dust particle?

ANYBODY!!!!
Thanks,
-Nolan

 

[nmmiles]

ttt

 

[mtn_extreme]

Do your own homework

 

[nmmiles]

lol, I cant. This one is really stumping me.

 

[adrenaline junkey]

The normal everyday weight of the dust particle in the earth's gravitational acceleration is given by
W = m2*g where m2 is the mass of the dust particle and g is 9.8 m/s^2.

The force of attraction of the dust particle to the lead sphere is given by
F = G*m1*m2/r^2
where G = 6.67*10^-11 N*m^2/kg^2, m1 is the mass of the lead sphere, m2 is the mass of the dust particle, and r is the distance between the particle and the center of the sphere.

So the ratio R is

R = (G*m1*m2/r^2) / m2*g =
(6.67*10^-11 N*m^2/kg^2 * 5600 kg * m2 / 0.3^2 m^2) / m2 * 9.8 m/s^2
(6.67*10^-11 * 5600 / (.09 * 9.8) N s^2/ m kg
The units all cancel out, as they should. (I find that watching the units helps me avoid errors.) So
R = 6.67*10^-11 * 6.350*10^3 = 42.3*10^-8 = 4.23*10^-7

 

[opus]

What ^^^ said. :/

 

[niner49]

Zero!

The sphere exerts no Gravitational FORCE on the dust particle, so the ratio would undefined.

 

[niner49]

The normal everyday weight of the dust particle in the earth's gravitational acceleration is given by
W = m2*g where m2 is the mass of the dust particle and g is 9.8 m/s^2.

The force of attraction of the dust particle to the lead sphere is given by
F = G*m1*m2/r^2
where G = 6.67*10^-11 N*m^2/kg^2, m1 is the mass of the lead sphere, m2 is the mass of the dust particle, and r is the distance between the particle and the center of the sphere.

So the ratio R is

R = (G*m1*m2/r^2) / m2*g =
(6.67*10^-11 N*m^2/kg^2 * 5600 kg * m2 / 0.3^2 m^2) / m2 * 9.8 m/s^2
(6.67*10^-11 * 5600 / (.09 * 9.8) N s^2/ m kg
The units all cancel out, as they should. (I find that watching the units helps me avoid errors.) So
R = 6.67*10^-11 * 6.350*10^3 = 42.3*10^-8 = 4.23*10^-7

Nice cut and paste AJ but you should have changed the numbers in the question on google to the one Nolan asked and then read the question asked about force versus mass versus acceleration

The sphere does not exert Gravitational pull on the particle.

Thanks for playing though

 

[nmmiles]

the shpere does exert gravitational force. Figured it out. the answer is 1.21*10^-7

Thanks for helpin tho.

 

[onlyrotax]

All I know is that I was always told "The heat of the meat determines the angle of the dangle".

 

[adrenaline junkey]

Nice cut and paste AJ but you should have changed the numbers in the question on google to the one Nolan asked and then read the question asked about force versus mass versus acceleration

The sphere does not exert Gravitational pull on the particle.

Thanks for playing though

Lmao well ya cant blame a guy for trying I had to spell check physics this stuff is so far beyond me

 

[skidoorulz]

The normal everyday weight of the dust particle in the earth's gravitational acceleration is given by
W = m2*g where m2 is the mass of the dust particle and g is 9.8 m/s^2.

The force of attraction of the dust particle to the lead sphere is given by
F = G*m1*m2/r^2
where G = 6.67*10^-11 N*m^2/kg^2, m1 is the mass of the lead sphere, m2 is the mass of the dust particle, and r is the distance between the particle and the center of the sphere.

So the ratio R is

R = (G*m1*m2/r^2) / m2*g =
(6.67*10^-11 N*m^2/kg^2 * 5600 kg * m2 / 0.3^2 m^2) / m2 * 9.8 m/s^2
(6.67*10^-11 * 5600 / (.09 * 9.8) N s^2/ m kg
The units all cancel out, as they should. (I find that watching the units helps me avoid errors.) So
R = 6.67*10^-11 * 6.350*10^3 = 42.3*10^-8 = 4.23*10^-7

Do you really expect us to believe that?

 

[Jeff C]

You wrote:

"R = 6.67*10^-11 * 6.350*10^3 = 42.3*10^-8 = 4.23*10^-7"

It is actually: R = 6.67*10^-11 * 6.350*10^3 = 42.3*10^-8 = 4.24*10^-7


But thanks for playing

 

[Wade]

A 1.10 -diameter lead sphere has a mass of 5400 kg. A dust particle rests on the surface of the sphere, which itself lies on the surface of the earth.

What is the ratio of the gravitational force of the sphere on the dust particle to the weight of the dust particle?

ANYBODY!!!!
Thanks,
-Nolan

Your question is weird. So "A 1.10 -diameter lead sphere has a mass of 5400kg."

1.10 what? 1.10 meter diameter sphere of lead weights 7665kg. ????

junkey did a pretty good job. But the question in essence is what is the ratio of the gravitational attraction of some unknown weight and size dust particle, to the planet versus a 5400kg ball, probably 1.10 meters in diameter.

F1 = G * m1 * m3 / r1^2
F2 = G * m2 * m3 / r2^2

m1 = mass of sphere
m2 = mass of earth
m3 = mass of dust
r1 = radius of sphere
r2 = radius of earth
G = is a constant
the radius of the dust particle drops out, too small.

So, F1 ratio F2
G * m1 * m3 / r1^2 ratio G * m2 * m3 / r2^2
G cancels, m3 even cancels and it's very small.

m1 / r1^2 ratio m2 / r2^2

5400kg / 0.55 m^2 : 5.9736 x10^24 kg / 637100 m^2

17851 : 1.47x10^13
1 : 824438716

 

[Dogmeat]

A 1.10 -diameter lead sphere has a mass of 5400 kg. A dust particle rests on the surface of the sphere, which itself lies on the surface of the earth.

What is the ratio of the gravitational force of the sphere on the dust particle to the weight of the dust particle?

ANYBODY!!!!
Thanks,
-Nolan

Whatever the quantity is, it's much much less than the weight of the dust particle ...

Right?

 

[triple]

All I know is that I was always told "The heat of the meat determines the angle of the dangle".

So is that your answer???

 

[Ex-Member]

Zero!

The sphere exerts no Gravitational FORCE on the dust particle, so the ratio would undefined.

Wrong, anything with mass exerts a gravitational force.

There's even an attraction between you and the ugly chick in high school.

 

[Wade]

Wrong, anything with mass exerts a gravitational force.

There's even an attraction between you and the ugly chick in high school.

Yha, be careful, or your two bodies will approach close enough to form an irresistible gravitational attraction. And, you'll fall into orbit around a superior body. Eventually, becoming tidal locked, and never being able to look at anyone but her.

 

[nmmiles]

yes, the 1.1 was in meters. I appologize, when I copied and pasted none of the units followed.

Yep, everything has a gravitational force. You exert a force on the earth just as the earth exerts a force on you. So that means when you jump you actually push the earth away from you and when you are up you are actually pulling the earth up towards you. Because the earth is so large and you are so small, the overall effects is so close to zero its not worth calculating.
Physics suxs....

 

[Wade]

yes, the 1.1 was in meters. I appologize, when I copied and pasted none of the units followed.

Yep, everything has a gravitational force. You exert a force on the earth just as the earth exerts a force on you. So that means when you jump you actually push the earth away from you and when you are up you are actually pulling the earth up towards you. Because the earth is so large and you are so small, the overall effects is so close to zero its not worth calculating.
Physics suxs....

Physics is cool.

You know, gravity is a funny thing. Their still not sure if gravity has properties of a wave, or if it's effects are faster than light. And, it's so weak, a whole planet, can only pull you too the ground at one G. Compare that to a little magnet, that can pick you off the ground. I think it's the weakest force known.