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Author Topic: So I was in the shower today...  (Read 20356 times)
Agamemnus
x/ \z
*****
Posts: 3491



« Reply #105 on: June 16, 2006, 02:45:12 AM »

In other news, I made an approximation of the gravity using my formulas and the result shows gravity first increasing and then sharply decreasing as the point of measurement reaches a depth at the center of the earth.

Code:

defint A-Z
screen 9
DIM C as double, F1 as double, F2 as double, intF1 as double, intF2 as double, F as double
DIM bigR as double, r as double, x as double
dim g as double, pi as double


G = 6.67300 * 10^-11 'm^3 kg^-1 s^-2
pi = 3.14159265
r = 63.78 'hundreds of kilometers
m1 = 5.98 * 10^24 'kg
C = G*pi*r*m1
'C = 10: r = 1000

for n = 0 to r

F1 = 0: for x = 0 to n
bigR = (r^2 - x^2)^.5
if bigR <> 0 then
intF1 = 0
for b = 0 to bigR
if n^2+b^2 <> 0 then
intF1 = intF1 + b/(n^2+b^2)
end if
F1 = F1 + intF1
next b
end if
next x

F2 = 0: for x = 0 to 2*r-n
bigR = (r^2 - x^2)^.5
if bigR <> 0 then
intF2 = 0
for b = 0 to bigR
if (2*r-n)^2+b^2 <> 0 then
intF2 = intF2 + b/((2*r-n)^2+b^2)
end if
F2 = F2 + intF2
next b
end if
next x
F = F1 - F2

if n > 1 then
'print F2;F;F1
'line ((n-1)*2+50, 300-log(abs(lastF))*10)-(n*2+50, 300-log(abs(F))*10), 15
line ((n-1)*2+50, 300-abs(lastF)/5)-(n*2+50, 300-abs(F)/5), 1
line ((n-1)*2+50, 300-abs(lastF1)/5)-(n*2+50, 300-abs(F1)/5), 2
line ((n-1)*2+50, 300-abs(lastF2)/5)-(n*2+50, 300-abs(F2)/5), 4
line ((n-1)*2+50, 300)-(n*2+50, 300), 15
end if
lastF2 = F2
lastF1 = F1
lastF = F
next n
'print abs(F)

sleep
system
Logged

Peace cannot be obtained without war. Why? If there is already peace, it is unnecessary for war. If there is no peace, there is already war."

Visit www.neobasic.net to see rubbish in all its finest.
Zack
*/-\*
*****
Posts: 3974



WWW
« Reply #106 on: June 16, 2006, 01:46:42 PM »

Quote from: "Agamemnus"
SJ Zero, you're making the same ridiculous assertions about what I said as before. And, if you didn't realize, that was a definite integral that actually made sense, at least to me, which I wasn't sure how to solve. Just because your integration notation is incomprehensible doesn't mean you need to trash talk me... and also you had no sources to back up your result.

Zack, the rate of acceleration remains constant, and so does acceleration.

a is acceleration in "a = 1/C".

Let's say m1 = 5:

a = 1/C
da/dt = 0 (da/dt is the rate of acceleration over time)

Let's say m1 = 10:

a = 1/C (the same...)
da/dt = 0


As you're not yet quite as hopeless as SJ Zero, Zack, I'll dissect your statement:

>In a seperate law, F=ma, and by extension a=F/m, acceleration of an object=force applied/the mass of the object, as the mass increases, the rate of acceleration DECREASES.
Yes, that is true, assuming the force is held CONSTANT.

>So as mass goes up, gravitational force does too
Also true, assuming the acceleration DECREASES.

>but we NEED more force to move a more massive object
Yes, that is correct, assuming acceleration stays constant.

>therefore things of different masses fall at the same speed.
I assume you mean gravitational acceleration here. That would be true, if you somehow reduced something else proportionately-- for instance, if you reduced the mass of the other object, or increased the distance.

Maybe I'm being misleading. Suppose we have two objects, one which is 5 kg and the other which is 10kg. To accelerate the 5kg object to 10 m/s, we need how much force? F=ma, F=5kg*10m/s, F=50 N. We need 50 Newtons of force. To accelerate a 10kg object to 10 m/s, we need double the force, because the mass doubled. We need 100 N. Suppose that we are getting our force from gravity. Like any other force, we need double to accelerate the 10kg object. Luckily, the force of gravity increases whenever the mass goes up. So 10kg is being pulled on more than 5kg. But we need more force to pull the 10kg object. So, the conclusion is that the 5kg and 10kg object will be moved at the same rate.
Logged

f only life let you press CTRL-Z.
--------------------------------------
Freebasic is like QB, except it doesn't suck.
Agamemnus
x/ \z
*****
Posts: 3491



« Reply #107 on: June 16, 2006, 03:29:51 PM »

>Suppose that we are getting our force from gravity. Like any other force, we need double to accelerate the 10kg object.
OK.

>Luckily, the force of gravity increases whenever the mass goes up. So 10kg is being pulled on more than 5kg.
Yes...

>But we need more force to pull the 10kg object.
We need more force to push a 10kg object, not pull it...

>So, the conclusion is that the 5kg and 10kg object will be moved at the same rate.
Now you're talking about velocity.. speed. v = d/t. So you're saying that the masses will pull towards each other at the same speed regardless of the mass, which is completely different than "as the mass increases, the rate of acceleration DECREASES"...

Your assertion is: the rate at which one object approaches the other object due to gravity is independent of the first object's mass.

Code:

v = d/t, a = v/t

F1 = m1*a1

F = G*m1*m2/d^2

Now suppose that F = F1, then you would be correct, I suppose.
F = F1 = m1*a1 = G*m1*m2/d^2 , so: v1 = t * G*m2/d^2

Can you say that F1 = F? Obviously if you do this you will be confusing the frame of reference. If F1 = F, then F2 = F, and F1=F2:
F1 = F2
m1*a1 = m2*a2

But we know that a1 = a2 (and v1=v2), and to say that m1=m2 is ridiculous. Ifthe point of reference was simply the object in question (e.g. F1=F2=F), NOT the other object:

1) The first object's velocity towards its own position is not dependant on the mass of the first object, but rather the second object's mass.

2) The second object's velocity towards its own position is not dependant on the second object's mass, but rather the first object's mass.
Logged

Peace cannot be obtained without war. Why? If there is already peace, it is unnecessary for war. If there is no peace, there is already war."

Visit www.neobasic.net to see rubbish in all its finest.
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