Gravitation

1.What's the weight of the body at the centre of the earth and why?
2.Why there is more probability of getting injured when we fall from the tall buildings than the shorter ones?

Reply 1

Varnick

1. A body has no weight at the centre of the earth, because it is pulled equally in all directions. (The ground above you cancels out the ground below you, etc.)

2. By newtonian mechanics, v^2 = u^2 + 2as, where s is your height above the ground, as a and u are (approximately) constant, the higher you are, the faster, and therefore harder, you hit the ground.

Reply 2

roshanhero

OP

9

For second one,
Can I say like this that as higher I fall from,with more velocity I will strike the ground.

Reply 3

roshanhero

OP

9

I still couldnot understand the first one mate.

Reply 4

Ans214

7

Yes, the higher you fall from, the faster you will be when you hit the ground (unless you've reached terminal velocity which is your 'top speed' due to air resistance. I dont think any building is tall enough to reach terminal velocity though).

For the first one, as varnick said, all the forces acting on you cancel out so you will have a zero weight. imagine a paper-clip between two magnets. it wont be pulled to either one because the forces acting on it toward each magnet cancels out.

Reply 5

bourney

12

For the second one though, when you get past terminal velocity, it doesnt matter how high the building you fall is from, your never going to go any faster. Youd just more likely suffucate.

Reply 6

G. Lee

8

roshanhero

I still couldnot understand the first one mate.

Mm... I would have gave a different answer for both. The original replyer said that:

1. A body has no weight at the centre of the earth, because it is pulled equally in all directions. (The ground above you cancels out the ground below you, etc.)

Which is partially true. We don't actually know if there is no weight at all. We have never experimented that far, however, it does go to say that as you radially go outwards from any source mass, there seems to be more gravitational force as you expand forward. It would be best to say you are ''tugged'' from all four corners of your mass in the center of a gravitational field, than saying that there is no weight at all. Instead, i would suggest your ''weight'' is ''tugged'' in equal directions as to almost be negligable so can be thoguht of being cancelled out. I would have also reserved a different answer for his second reply:

2. By newtonian mechanics, v^2 = u^2 + 2as, where s is your height above the ground, as a and u are (approximately) constant, the higher you are, the faster, and therefore harder, you hit the ground.

Thus is not the same equation i would have provided as the answer. I would have used the potential equation:

-\frac{h}{2}=\frac{v*^2-v^2}{-2g}

And thus, this equation works for velocities of the said object (in this case yourself) to be zero so that

\frac{h}{2}

means that you are these many units above the ground. You can then measure the probability as being very slim of survival as the height gets further up, for, when you jump off a building, you are a accelerating due to gravity, so that you may end up with

\frac{4}{3} gh = v*^2

since v=0. This is the basics of work and potentiality.

Reply 7

Kyle_S-C

12

roshanhero

I still couldnot understand the first one mate.

Essentially you want to work out the net gravitational force on the body at the centre of the Earth.

First make the approximation that the Earth is perfectly spherical and you somehow exist at the centre (in some kind of spherical hole if you want to be fussy :p:

). Also note that the Earth is the only body which acts on you (since we're considering the force from the Earth).

Now, consider the net upwards force from the hemisphere above you and the hemisphere below you. The masses should be equal and the directions are opposite, so doing a vector addition the forces cancel out (one pulls you up the other pulls you down with the same force).

Since the Earth is approximated to be spherical, the situation is the same whichever direction you face in, which means that there's no net force in any direction, so you have zero weight.

I think the same is true when you're at the centre of mass of any system, it should exert no gravitational force upon you. In the real Earth the centre of mass will be slightly off centre due to mountains and oceans and stuff, but I can't imagine it's that bad an approximation to make it spherical.

Reply 8

Kyle_S-C

12

G. Lee

2. By newtonian mechanics, v^2 = u^2 + 2as, where s is your height above the ground, as a and u are (approximately) constant, the higher you are, the faster, and therefore harder, you hit the ground.

Thus is not the same equation i would have provided as the answer. I would have used the potential equation:

-\frac{h}{2}=\frac{v*^2-v^2}{-2g}

That equation is very similar to a rearrangement of v^2 = u^2 + 2as (h = s, g = a etc.), but you're missing a factor of 2...the dimensionality is right, I'm an idiot.

Reply 9

G. Lee

8

Kyle_S-C

That equation is very similar to a rearrangement of v^2 = u^2 + 2as (h = s, g = a etc.), but you're missing a factor of 2...the dimensionality is right, I'm an idiot.

Look my friend, you are not an idiot.

There are two people i trust in the world.

One of them is not myself, and the other is anyone but God... but what is trusted here, is the quintessential nature of mankind... the inescapable ability to make mistakes. We are not perfect, and just because you did not realize ''out of sight'' straight away that the equations where actually dimensonsionally-correct no more makes you an idiot, than i am a developing scientist, because trust me, i've probably made more mistakes mathematically in my life than yourself friend.

So do not batter yourself up in cold cooking oil.

(that last saying was mine... F*** knows where it came from... :biggrin: