# Equivalence Principle [closed]

Let’s imagine that the Equivalence Principle did not exist. Let’s assume that gravity works like any other force. That greater masses fell at a faster rate than smaller masses. Under this new theoretical principle, how much faster would we expect a cannon ball to drop than a feather given the difference in the measurable masses (lets say the feather weighs 0.000001 kilo grams and the cannon ball weighs 10 kilo grams )

Could this be calculated and if so are they falling at a visibly different rate or is the rate still extremely small ?

## closed as unclear what you're asking by sammy gerbil, stafusa, Kyle Kanos, Jon Custer, Chris♦Jan 31 '18 at 0:58

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• Why they wouldn't fall at the same rate? I think it would not make sense – L.Gyula Jan 29 '18 at 17:05
• Gravity does work like any other force. – Jon Custer Jan 30 '18 at 14:56

Let’s assume that [...] greater masses fell at a faster rate than smaller masses.

In this case, I think that what you really want to do is to redefine the gravitational force in a new way, such that the acceleration given to a falling body by this force is mass-dependent. Let's try the simplest way to add a mass dependence in the gravitational acceleration: we can define $g'(m)$ such that $$g'(m)=g_0+k\cdot m,$$ where $g_0$ and $k$ are constant while $m$ is the mass of your body. In the real case of course is \begin{cases} g_0=9.8\,m/s^2\\ k=0\,m/kg\cdot s^2. \end{cases}

Given this form of $g'(m)$, one could simply calculate the time taken by a body to hit the ground by the formula $$s(t)=s_0+v_0\cdot t-\frac{1}{2}g'(m)\cdot t^2.$$

Please note that the above formula for $g'(m)$ contains a linear dependence on $m$ because that's the simplest dependence I can think about, but one can play with that adding different terms.

How much faster would we expect a cannon ball to drop than a feather?

This really depends on what is the value of $k$ in the formula written above: the time difference could be negligible for small value of $k$ (note that $k=0$ in the reality), or very relevant for greater value of $k$.

The equivalence principle is not that unusual, consider dropping two identical cannon balls. Of course they fall at the same speed. Now place both balls in a bag and they still fall at the same speed. Nothing unusual about that. The same goes for Everything else. Drop two identical atoms and they fall at the same speed, now fill a cannon ball with the same type of atoms and together they too fall at the same speed. When you think about it there is no reason multiple masses should fall faster than one and the equivalence principle is just a given.

• But if we decide to drop two objects of differing weights I.e a cannon ball and a feather then instinctively we feel they might fall at different rates, I believe this is what got Einstein’s juices running in his early pursuit of relativity. The question I feel is not fully answered is surely the mass of the feather and the canon ball is relative to the other gravitational force acting on them which is the earth. In which case both the feather and cannon ball will be affected alike because there relative mass to the earth is negligible – Harvey Jul 6 '18 at 0:02
• No, the feather runs into air resistants so you cannot compare. On the moon where there is no atmosphere the feather would fall at the same speed. – Bill Alsept Jul 6 '18 at 0:06

Sure, that'd just be saying that inertial mass is different than gravitational mass

$m_i a = m_g g$

So that masses wouldn't be the same, they wouldn't cancel out anymore, and $a$ would be no longer $g$, but

$a= \frac{m_g}{m_i}g$.

This would give you the acceleration. Now you'd have to integrate this acceleration to get velocity and position.

If $g$ is constant it would just be the typical Uniformly Varied Straight Movement.

$y_F=y_0 + v_ot - at^2/2$

But $a$ wouldn't be $g$ now, but the upper expression.

Note:

as I say in my comment, I'm supposing that, asthe equivalence principle says that an accelerated observer is indistinguishable from being in a gravitational field, and that's because $m_g=m_i$; if they aren't the same anymore, each object will fall at a different rate, a rate given by their own ratio $m_g/m_i$. I am assuming that $m_g$ is not the same as $m_i$ now, nor it is the same ratio for all bodies. If the ratio were the same, we would still have $a\propto g$, so we would only need to redefine $G$ so that the gravitywere actually the same as $a$.

However, if the equivalence principle doesn't hold, each body should have a different ratio, and so they'd fall with different accelerations.

• The answer is incorrect. The proportional difference between inertial and gravitational masses you are describing is accounted for by the value of the gravitational constant. For these two masses to be really different they must not be proportional. Only then the Newtonian gravity would fail. Otherwise all things still fall at the same rate. Besides it is not clear what A is, what UVSM stands for, plus spelling errors and lack of clarity on what your conclusion is at the end. – safesphere Jan 29 '18 at 16:34
• I already corrected my mistakes, it's true: I type too badly, but it was still understandable. Now, as for the question, I am assuming that $g$ is constant, following the example of a constant electric field, which causes an acceleration given by $qE/m$. If I'm not wrong, the equivalence principle says that an accelerated observer is indistinguishable from being in a gravitational field, and that's because $m_g=m_i$. If they aren't the same anymore, each object will fall at a different rate, a rate given by their own ratio $m_g/m_i$, which needn't be the same for all bodies. – FGSUZ Jan 29 '18 at 22:49
• It seems critical for your answer, but not obvious, that $m_g/m_i$ is not the same for all bodies. I would suggest making this a part of your answer. – safesphere Jan 29 '18 at 23:51