# Mountain and a feather dropped from same height. Which hits Earth first? [duplicate]

Let's assume:

• the feather has a mass of $0.006$kg
• the mountain has a mass of $6 \times 10^{15}$kg
• the earth has a mass of $6 \times 10^{24}$kg
• no air resistance
• both objects released at the same time with an initial velocity of $0$
• both objects released at a height of $10,000$m

Would the mountain hit first, as it itself would exert a gravitational pull on the earth?

• For the miniscule effects the answerer will be looking at, it is important to know how the mountain and feather start at 10000m - head to head, or mid to mid, or end to end, or differently? Also, no air resistance - but Newtonian gravity or general relativity? (I'm sure I can think of further nitpicks with time...) – ACuriousMind Jan 5 '15 at 0:32
• Good question; possible duplicate of physics.stackexchange.com/q/3534 and physics.stackexchange.com/q/156634, which was asked perhaps 5 minutes before yours. – HDE 226868 Jan 5 '15 at 0:47
• The question as it is, I suspect isn't asking the question you really want to know. Which is, in TWO SEPARATE experiments, two objects, a mountain vs a feather are dropped. Which hits earth first. In that case the answer is the mountain. – Aron Jan 5 '15 at 6:38
• Agreed, the intention is to test the two independently. We don't want to start worrying about the feather being attracted to the mountain! – AndyM Jan 5 '15 at 19:21

TL;DR: If the race is made interesting, it's the feather that hits first. Otherwise, it's the mountain.

Simplifying assumptions:

I'll make some spherical cow simplifying assumptions. Specifically, the Earth, the mountain, and the feather are spheres. The spherical mountain, being made of typical rock, has a density of $\frac {1125}{128\pi}\,\text{g/cm}^3\approx 2.8\, \text{g/cm}^3$. This nicely makes the spherical mountain have a radius of 8 km. The feather, being made of ultralight feather stuff, has a density of $\frac 9 {16\pi}\,\text{g/cm}^3\approx 0.18 \,\text{g/cm}^3$. This nicely makes the spherical feather have a radius of 2 cm.

The other assumption is the meaning of "both objects released at a height of 10,000 m." I see three possible interpretations of height: measured from the surface of the Earth to the top of the object, from the surface of the Earth to the center of the object, or from the surface of the Earth to the bottom of the object.

The first interpretation (height is measured from the surface of the Earth to the top of the object) doesn't make any sense given the circumstances. If the top of the mountain is 10 km above the Earth, the bottom of the mountain is 6 km below the surface. There is no race; the mountain has already won.

The second interpretation (height is measured from the surface of the Earth to the center of the object) also doesn't make much sense given the circumstances. If the center of the mountain is ten km above the Earth, the mountain only has to fall two km before it hits the Earth. The feather on the other hand has to two centimeters shy of ten km. This gives the mountain an eight kilometer advantage. This isn't even close to an interesting race; the mountain falls two kilometers the Earth in less than half the time it takes the feather to fall nearly ten kilometers.

The third interpretation (height is measured from the surface of the Earth to the bottom of the object) is the only one that makes for an interesting race. At the start, the feather's center of mass is 10.00002 km above the surface of the Earth while the mountain's center of mass is 18 km above the surface of the Earth. This means the feather's acceleration due to Earth gravity is slightly more than that of the mountain. This gives the feather a slight advantage over the mountain.

To make this a one dimensional problem, we'll drill a 4 cm deep, 4 cm diameter cylindrical hole in the mountain. The spherical feather will just fit in this hole. I'll ignore that this hole makes the mountain not quite spherical.

Who wins?

Now let's consider two races. In the first, we'll place the feather in that little pocket in the mountain, orient the mountain so the pocket faces downward, and release the spherical mountain and spherical feather simultaneously. The winner is the object that first reaches the surface of the Earth. In the second, we'll release the spherical mountain and spherical feather separately and time the separate falls. The winner is the object that takes the lesser amount of time to reach the surface of the Earth.

In the first race, the feather accelerates downward toward the Earth and upward toward the mountain. The mountain accelerates downward toward the Earth and downward toward the feather. The Earth as a whole accelerates upward toward both the mountain and the feather. Denoting $h_f$ as the height in kilometers of the bottom of the spherical feather above the surface of the Earth, $h_m$ as the height in kilometers of the bottom of the spherical mountain above the surface of the Earth, $x$ as the position in kilometers of the center of the Earth relative to it's initial position, $r$ as the radius of the Earth in kilometers, and $g=GM/r^2$ as Earth surface gravity, the differential equations that govern the first race are

\begin{aligned} \ddot h_f &= -g\left( \left( \frac r {r - x + h_f + 0.00002} \right)^2 - 10^{-9} \left( \frac r {h_m - h_f + 7.99998} \right)^2 \right) \\ \ddot h_m &= -g\left( \left( \frac r {r - x + h_m + 8} \right)^2 + 10^{-27} \left( \frac r {h_m - h_f + 7.99998} \right)^2 \right) \\ \ddot x &= g\left( \left( 10^{-9} \frac r {r - x + h_m + 8} \right)^2 + 10^{-27} \left( \frac r {r - x + h_f + 0.00002} \right)^2 \right) \end{aligned}

Integrating numerically, the feather wins by about 0.042 seconds. The qualitative explanation is that the slight upward acceleration of the feather toward the mountain does not make up for the greater downward acceleration the feather experiences toward the Earth compared to that of the mountain. The acceleration of the Earth as a whole toward the mountain and the feather is irrelevant in this race.

What about the second race? In this case, the acceleration of the Earth as a whole toward the mountain might well be relevant. The acceleration of the Earth as a whole toward the feather is still irrelevant because the feather's mass is so tiny. It turns out that there is essentially no difference in timing. The feather has to fall an extra 10 micrometers than does the mountain. The feather still wins by 42 hundredths of a second.

What if we use point masses?

In this case, the first race is a tie, but the mountain wins the second race.

I am sorry to say but most of the other answers given are wrong. The correct solution is to use the reduced mass: http://en.wikipedia.org/wiki/Reduced_mass. The approximation that two bodies are falling with the same acceleration in the gravitational field of a third is only valid for light test masses (that's the equivalence principle). If we have to look at the dynamics of heavy objects gravitating around each other, the accelerations in the center of mass system are given by the reduced masses. That, too, is a consequence of relativity, after all, the two masses are perfectly equivalent with respect to gravity.

• Since the mass of the mountain is $10^{-9}$ times the mass of the Earth, the reduced mass is slightly less ($10^{-9}$) than the mass of the lighter object. If you work through the math, you will probably notice that you get the same answer as Newton would (with a little error). – LDC3 Jan 5 '15 at 0:54
• @LDC3: The question was specifically if there will be a difference. The answer is, YES, there will be a difference, it's not a null experiment. Reduced mass is physics 101. The link to the previous question has plenty of correct answers. The reduced mass IS what Newton would get if he did it right. – CuriousOne Jan 5 '15 at 0:58
• This is not answer. It doesn't say whether the feather or the mountain hits first. – David Hammen Jan 5 '15 at 13:49
• @DavidHammen: And while it's not an answer, it is correct. :-) – CuriousOne Jan 5 '15 at 13:54

The force that an object experience due to the attraction of a nearby object is: $$F=G\frac {m_1\times m_2}{r^2}$$ We also know that $F=ma$. Substituting the gravitaional constant times the mass of the Earth for another constant: $k=G\times m_1$. We have: $$F=m_2\times \frac {k}{r^2}$$ $$\frac {F}{m_2}=\frac {k}{r^2}$$ or $$a=\frac {k}{r^2}$$ This means that whatever the object, the acceleration depends on the gravitational constant, the mass of the earth and how far away the object is from the center of the earth. It does not depend on the mass of the object at all.

• @Aron The mass of the mountain is $10^{-9}$ times the mass of the earth, which creates an error that is too small to be detected. I also put this under CuriousOne's answer. But I do agree with him in that the mountain will contact first. – LDC3 Jan 5 '15 at 3:22