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I’m having a debate with my father. He says that it is perfectly understandable why objects fall with gravity at the same rate despite being of differing masses. His point is, in relation to the mass and gravitational pull of the Earth the difference between a feather and a cannon ball’s mass relative to the mass of the earth is negligible. Therefore gravity acts on them in the same way (or very nearly almost the same way). Is he right?

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The mass of the earth actually does not define which of the two bodies (feather or cannon ball) falls first. The problem is a bit more deep that that, and it is actually the reason behind General Relativity and in general, all geometric formulations of gravity.

Roughly speaking, the way masses move is solely dependent on the local properties of the gravitational field, independently of their mass. This observation lead Einstein to the formulation of the Equivalence Principle, and is the reason we think of gravitational force as a reflection of the local curvature of space time.

If you're familiar with a bit of math, if $m_{\rm grav}$ is the gravitational mass of a body (measure how much it is affected by gravity) and $m_{\rm inertial}$ is its inertial mass (measure how difficult is to change its state of motion), then Newton's second law states that when this body is the presence of a mass $M$ its acceleration follows the expression

$$ m_{\rm inertial}a = G\frac{m_{\rm grav}M}{r^2} $$

This provide a way of measuring the ratio in the lab

$$ \frac{m_{\rm inertial}}{m_{\rm grav}} = \frac{GM}{ar^2} $$

The result is that this number is one with an accuracy of 1 in 20 million. So we are tempted to believed that indeed inertial mass is the same a gravitational mass and consequently, that all masses are affected the same way by gravity, or said in other words, that gravity is an emergent property of the local geometry of spacetime.

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No, he is not right. The reason is that the gravitational mass of an object is equal to its inertial mass. For all we know this is exact and there is no approximation involved besides the assumption of no air resistance.

In more detail: The gravitational "pull" force with which the earth (of mass $M$) pulls on an object of (gravitational) mass $m_g$ at a distance $r$ from the center of earth, is given by: $$F=\frac{GMm_g}{r^2}$$

As you can see, this force depends on the mass of the object and will be much bigger for the cannon ball than for the feather. Just put a cannon ball on your head and you will feel the difference compared to having a feather on your head.

According to Newton's law this force causes the object to accelerate with acceleration: $$a=\frac{F}{m_i}$$ where $m_i$ is the inertial mass, which you can imagine as a resistance to acceleration.

Making use of the fact that $m_g=m_i$, you find from these two equations that the acceleration is independent of the object's mass and only depends on the distance from earth: $$a=\frac{GM}{r^2}$$ which on the surface of earth ($r$ equal to earth radius) is the familiar acceleration $g\approx 9.81~m/s^2$.


Or in words:

  1. Earth pulls stronger (larger force) on heavier objects
  2. Heavier objects are harder to move (require more force to accelerate)

Miraculously, for reasons unknown, it turns out that these two effects just cancel each other such that the acceleration is the same for heavier and lighter objects.

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Yes, he's right. The gravity well of the earth is significantly greater than both that of the feather and the cannon ball. As such, they travel along nearly identical geodesics at nearly identical rate.

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protected by Qmechanic May 12 '17 at 19:11

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