User talk:Myktk

Critique1 Suppose two masses m1 = m2 = 1 kg, Both masses are in space and the centre to center distance between them is 1 metre = r , then

F= Gm1m2/r^2 = G = 6.67*10^-11 Newton.

But Gravitational accelerations or forces of attractions of both masses balance each other and therefore value of gravitational force F should be zero between those two masses, if not, then

Which one is gravitating/ or falling mass. And how aforementioned masses attract each other by F= Gm1m2/r^2 = G = 6.67*10^-11 Newton, when there is no gravitating or falling mass.

Critique2 Make it more simple let say earth is an sphere. Now if we put imaginary earth on earth or sun on sun and now apply F= GMm/R^2 to the masses, where g = GM/R^2. M= mass of earth, m= mass of imaginary earth, centre to centre distance between two masses = diameter of earth. Now again which one is falling mass or gravitating mass where gravitational acceleration of each mass cancel each other.

Critique3 Suppose there are two spheres mass of 1st sphere = m1, diameter of 1st sphere = d1= 1metre mass of 2nd sphere = m2, diameter of 2nd sphere = d2= 1 kilo metre are at the same height (say 10 metres) from the surface of moon

Would they truly fall towards earth with same acceleration in absence of air resistance on the surface of moon. I say no. Reason Radius of moon = 1,738 km So centre to centre distance between m1 and moon = R1 = 1738.001 km and m2 and moon = R2 = 1739 km Now apply F=GMm/R^2 to each mass , where M = mass of moon and (g=GM/R^2) B/w moon and m1 F=GMm1/R1^2 , g1 = GM/R1^2 And b/w moon and m2 F=GMm2/R2^2, g2 = GM/R2^2 Please also note: if we dropped aforementioned masses on the surface of moon at the same rate by removing a wooden plank (say) from their bottom. Then application of F= GMm/r^2 to each mass will show different results because of the difference in values of R1 and R2.

Critique4 Here is the summary of attraction Forces of Sun - Moon, Earth - Moon and Sun - Earth during Total Eclipse when the Moon is between Sun and Earth.

F = GMm/r^2

Sun ----------------------------------- Moon ------------ Earth S-M = 4.1984 x 10^20 M - E = 2.2 x 10^20 Net force on the Moon = 4.1984 x 10^20 Minus 2.2 x 10^20 = 1.998 x 10^20 towards Sun At this point why Earth force the moon to revolve around its centre when the net force on the moon is much much greater towards the sun ? Explain please. Check the calculation pls. If we consider the Sun - Earth Force S - E= 3.67 x 10^22 , then Net Force on the Moon = 1.998 x 10^20 Plus 3.67 x 10^22 = 3.68 x 10^22 towards Sun.

Please also note that force of attraction F = GMm/r^2, between sun and moon in any case (perigee, apogee, average) is much greater than between moon and earth F = GMm/r^2, (perigee, apogee, average) . So technically it should revolve around the sun not earth. So why moon revolves around earth? 75.152.153.134 (talk) 05:01, 4 August 2008 (UTC)zarmewa

Answer to Critique1&2: There is only one gravitational force "pushing" or "pulling" objects together. Objects of the same mass do not "balance" or "cancel" the gravitational force between them. Which one is falling depends on your point of view. If you were on mass m1 you would say m2 is falling. If you were on m2 you would say m1 is falling. If you were some distance away from them you might say they were moving toward each other.

Answer to Critique3: The diameters of the masses does not matter. The formula for the force of gravity only uses there masses and distances apart. The force of gravity is f=g*m1*m2/r^2, where f=force of gravity, g is gravimetric constant, m1 is mass of the moon in this example and m2 is mass of the falling object and r is the distance from the centers of the moon and each object and is squared in this formula. The formula for acceleration, or speed they would fall, is a=f/m. Where a=acceleration, f=force of gravity and m=mass of object. Assume two objects, object A is 1kg and object B is 100kg and both were dropped on the moon from the same height, 10 meters. For object A, f=g*moon mass*1kg/(moon radius+10 meters)^2. For object B, f=g*moon mass*100kg/(moon radius+10 meters)^2. Since everything is the same except for the mass of the two objects fA*100=fB. Or since object B is 100 times as massive, force of gravity is 100 times as strong, which is why it would weight 100 times a much. So acceleration for object A is aA=fA/massA and for object B is aB=fB/massB or aB=fA*100/massA*100 or aA=aB. So the ecceleration for both objects is the same. Both would fall at exactly the same speed. The same would be true on the earth without air resistance. They would fall faster on the earth than on the moon though since the earth is more massive than the moon.

Answer to Critique4: The moon does in fact orbit the Sun. It also orbits the earth. If you were to draw a large circle to represent the orbit of the earth, and then label the months evenly around the circle, then draw the moon for each of its four phases for each month, it can be seen that the moon orbits the sun in a wavy circle as it is orbiting the earth. JohnXR (talk) 06:42, 3 February 2009 (UTC)

Censures to your

Answer to Critique 1&2 and 3:

Here is Newton’s equation, F=GMm/R^2

M=mass of earth m=mass of 1kg on the surface of earth at rest (remember not moving) R= radius of earth (radius of small mass on earth is neglected) G= Universal constant is used in order to get the value of “g” for equation f=ma=mg.

Now increase the size of aforementioned mass “m” to imaginary earth and again, apply the Newton’s law

F=GM^2/D^2 where D=diameter of earth viz c/c distance b/t two masses.

Thus if gravitational law is true then GM/D^2=g=acceleration due to gravity , is true. What do you think value of G can justify the value of “g” at that height.

Further is it possible for a body to have acceleration “g” at rest position? If not how f=mg is possible for a body is at rest position. Similarly acceleration or “g”depend upon time and time dilate as per Sir Eienstein as one away from celestial bodies. For “g” time has to be constant thus if Eienstein is right then Sir newton is wrong. Thus if gravity is wrong how could Sir Eienstein talk about black hole gravity and the same is apply to neutron star.

2nd part: As m1 attract m2 due to its gravity and m2 attract m1 due to its gravity. Both gravitational forces are same in magnitude but opposite in direction. How could they move towards each other. Similarly in order to satisfy newton’s gravitational law one has to be falling and the other gravitating mass and how value of G has to be adjusted in the equation. Since both masses are staying at their position therefore how come a force of value G newton is exits. This means value of universal constant G is wrong which fail to justify the Newton’s law in aforementioned cases. Sir Eienstein used value of G in his many important equation.

g= GM/R^2, it does not depend upon the mass of the falling object. I just wanted to tell you that mass with bigger size will hit the ground first because of its bigger size.

Answer to Critique 4:

What do you think I don’t know about this. Grab a calculator, pencil and piece of blank paper and find all the required values and apply Newton law of gravitation to the following cases.

1: sun and moon in apogee, perigee and average (s-e-m, s-m-e, s-e when m at right angle to e approx both e and m are at same distance from s) 2: sun and earth in apogee, perigee and average and 3: earth and moon in apogee, perigee and average

s ----------e------m, s-------- m ----e, s-------------------e,m is up or down at right angle to e.

After finding all values of forces then do the simple math (+, -) for F. You will find gravitational force b/t sun and moon in any of above case is much greater than gravitational force b/t earth and moon. So if newton law of gravitation is true then moon should revolve in a separate orbit not along with earth around sun.

critique 5:

Let “P” is a point or an origin of two circles of radius r1=1 meter and r2= 2 meter. Consider these two circle as spheres (empty from inside) or consider these circles as two bangles in space. Now apply Newton’s law of gravitation i.e. F=GMm/R^2 to those two masses and neglect all other local attractions. As gravitational force of attraction between these two bangles is infinity (center to center distance b/t masses is zero) Now how much force is required to separate aforementioned masses, infinity or less? If less then what about the Newton’s law?

Critique 6:

Galileo had concluded hundreds of years before - All objects released together fall at the same rate regardless of mass. i.e. g = GM/R^2 and had proved on the lunar surface by falling feather and hammer at the same time in the absence of air.

Atoms and molecules of air or gases have also masses. Therefore there was no one with the brain who could ask Newton/Galileo that why the masses of atoms or molecules of air or gases do not fall at the same rate on the surface of earth along with other masses or obey the Newton law of gravitation. Further movements of air molecule in atonosphere depend upon temperature and Newton did not mention any temperature in his law of gravitation. —Preceding unsigned comment added by 96.52.178.55 (talk) 08:02, 20 February 2009 (UTC)