# Do all object really fall at the same rate regardless of mass?

#### pugilist777

Ive always had this question in the back of my mind? Do all objects fall at the same rate regardless of mass? This theory has been tested many times. I just watched a video on youtube where they dropped a bowling ball and a feather at the same time in a giant vacuum chamber. As predicted, they fell in perfect unison, AMAZING!
I have a hard time understanding this though. From what i have found, the reason for this is that more massive objects have a stronger attraction to each other due to gravity. They also have more inertia which balances out the force of gravity. Therefore causing objects to fall at the same rate.
So answer me this. Hypothetically speaking, if the earth were to completely stop in its orbit, would it fall into the sun at the same rate as a feather? Im not too educated on equations and such but thankfully there is online calculators for problems like this. According to this online free fall calculator https://www.omnicalculator.com/physics/free-fall it says that the amount of time it would take for an object to fall into the sun (force of gravity set at 28.02 times that of earth) (initial velocity set at 0) (height set at distance from earth to sun 149600000 km) would be 9.166 hours. Then i asked google how long it would take the earth to fall into the sun if it completely stopped in its orbit. The answer, about two months, or 65 days! Thats not even remotely close to 9.166 hours. I must be missing something. Whats going on here? Thank you for your time and insight!

#### ChipB

PHF Helper
The problem is that your figure of acceleration due to gravity of the sun is incorrect. The force of gravity due to the sun's mass acting on an object on the sun's surface is indeed 28 times greater than the force due to earth's gravity at the Earth's surface. But the Earth is not on the sun's surface (thankfully). The acceleration due to the sun's gravity at the distance of the Earth from the sun (about 93 million miles) is approximately 0.0059 m/s^s, which is about 1/1650 of the acceleration due to gravity on Earth. So it's going to take a lot longer than 9 hours.

By my quick calculation I get an answer of around 64 days, so Google seems to have it right.

3 people

#### pugilist777

Aha! That totally makes sense, thank you. So that free fall calculator i used is basically intended for objects falling short distances? Because the acceleration of gravity changes drastically over long distances vs falling A relatively short distance such as falling out of an airplane to earths surface?
So the fact that all objects fall at the same rate holds true to very large objects as well? So this means that a feather will fall to the sun at the same rate as the earth will?

#### ChipB

PHF Helper
Yes, in general massive objects fall at the same rate as low mass objects - as Galileo demonstrated all those years ago - assuming that air resistance can be ignored. One can think of situations where this might not be so - if the large object is very large - on the order of magnitude of the Earth's size (or the sun's, or whatever the object is falling towards), or if the gradient of the gravitational field is significant, then tidal forces can play an effect and things can get weird. But for every day objects like canon balls, feathers, etc. the acceleration of an object due to gravity is independent of its mass.

1 person

#### pugilist777

Okay its starting to make sense now. So what do you mean by tidal forces? And you say things get weird? Like gravity behaves differently than expected? Ive been researching about gravity a lot and from what i have read, gravity isnt necessarily universal? Einsteins relativity works at the macroscopic level. But it breaks down at the atomic level and at very large scales such as super massive black holes (correct me if im wrong). It just seems like everything should work under the same principles no matter the size. For example another question ive had for a while is light bending around corners (diffraction). Is this gravity that causes that? Very massive objects bend light, so technically smaller objects should too, just at a very small scale. The affects of gravity are also dependent on distance, so even though the gravity of a small object is very small, the light rays are passing by extremely close so therefore it should bend the light, right?

#### Woody

Tidal forces relate to the change in gravity with position.
For most circumstances (on the Earth) we assume gravity is constant, because the change is very small.
However the tides in the oceans are caused by the gravity from the moon changing (by the reciprocal of the distance squared) from one side of the Earth to the other.

In more extreme environments (black holes etc.) the tidal changes in gravity with position can be much larger.

Regarding very close approaches of light to less massive objects
I think that if you were to work out the math,
the tiny distances you would require to see any significant gravitational effect
would place the photons within literal touching distance of the objects

This would mean that you would have other (electromagnetic) interactions happening between the photons and the electrons of the object,
which would totally swamp any gravity effects.

#### pugilist777

Tidal forces?

So with very large objects such as earth and black holes, the acceleration of gravity would be faster/stronger than a feather falling to earth? In other words, would the earth fall faster to the sun than a feather would? Im not too clear on what you guys mean when things get weird when you start dealing with tidal forces. Thanks much!

#### topsquark

Forum Staff
So with very large objects such as earth and black holes, the acceleration of gravity would be faster/stronger than a feather falling to earth? In other words, would the earth fall faster to the sun than a feather would? Im not too clear on what you guys mean when things get weird when you start dealing with tidal forces. Thanks much!
Barring anything getting in the way any two objects will fall with the same acceleration due to gravity. For an object with mass m1 falling only under a gravitational force by m2:
$$\displaystyle F_g = \dfrac{G m_1 m_2}{r^2} = m_1 g$$

So $$\displaystyle g = \dfrac{G m_2}{r^2}$$

As the m1 has canceled out the accelertation will be the same for any value of the mass. This holds in GR as well.

-Dan