# What is acceleration due to gravity at the centre of the earth? [duplicate]

When I asked my teacher that what is the acceleration at the centre of earth, he replied that it is 0 as when we move inside the earth, the effective mass decreases i.e. the mass that exerts gravitational force on us decreases and hence at the centre of the acceleration due to gravity is zero.

However when we reach at the centre of earth the radius is also zero hence from the formula - GM/r^2 Where M is mass of earth and r is it's radius, by putting 0 in place of M and r , we get value of acceleration due to gravity = 0/0 which is an undefined quantity.

Even if my teacher was wrong, on Google, the answer for value of acceleration due to gravity is zero and the radius of earth at its centre is also zero so by the formula:- GM/r^2 putting 0 in place of r, we get GM/0 So we get GM/0=0 But GM/0 is again an undefined quantity.

I want ask how is acceleration due to gravity 0 when it is not permitted by the formula as the formula should also be satisfied in every case.

• Possible duplicate of Would you be weightless at the center of the Earth? Aug 28, 2018 at 18:34
• No my question is not a duplicate of that question as that question asks if the gravitational acceleration at the centre of earth is zero while from my question, u want to ask that WHY THE FORMULA DOES NOT PERMIT 0 ACCELERATION DUE TO GRAVITY AT THE CENTRE OF EARTH. Aug 29, 2018 at 3:20
• Both linked duplicates have an answer with the desired equation in it, no? Aug 29, 2018 at 9:56
• no there is no answer to my question i those links Aug 29, 2018 at 10:00
• Hi, welcome to Physics! Note that our homework-and-exercises tag describes questions that are "homework-like," regardless of whether they originated as part of some assignment; please don't remove the tag again.
– rob
Aug 29, 2018 at 19:44

The best way of approaching this (no pun intended) is to consider what happens when you get closer and closer to the centre.

When you are at distance $r$ you have $$m(r)=\left(\frac{4\pi \rho}{3}\right)r^3$$ kg of mass below you (remember, it is only the mass inside a radius that matters the gravity of spherical mass distributions - very convenient). It exerts an acceleration $$a(r)=\frac{Gm(r)}{r^2} = G\left(\frac{4\pi \rho}{3}\right)r.$$ Notice how this is well-behaved at $r=0$ - the acceleration just decreases linearly as $r$ decreases, and eventually approaches 0.

(The above formula is of course only valid from $r=0$ to the surface $R$ of the Earth, then $m(r)$ stops increasing and you get the normal acceleration formula $a(r)=Gm(R)/r^2$.)

The gravitational law only applies if you have a point mass at the center. If you have an extended body like the earth it is still valid if you have a sphere and you are not inside it.

If you now dig a hole, all the matter above you will attract you and cancel out some of the gravitational force from below. Mathematically you have to solve the integral of the small forces over all points of the shpere. I will skip the rigours calculation and give you the result straight away:

Inside the shpere effectivly only the part below you will attract you ( all other parts cancel out each other). So the further you dig to the center of the earth the smaller the gravitational force until it reaches zero on the center.

Imagine an air molecule in a balloon that is in the middle of space. (Not likely but bear with me) That air molecule is being pulled by all the pieces of that balloon, and it always wants to feel balanced. In this case it is going to find a spot where all the forces are pulling on it equally- the center. At this location it's not moving in any direction but being pulled on by every peice of the balloon. Now let's stick you in the center of the earth. You are being pulled by every piece of matter in the earth and this then balances you so you aren't moving. The only missing peice is that since we are three dimensional and have limbs each limb is being pulled by the earth above and below and to each side so no movement occurs.

• That was correct and thank you for that but please explain why using the formula , we get a different result. Aug 29, 2018 at 6:34

Newtonian gravity only works when you consider two (or more) objects some distance apart. You're trying to apply the formula for gravity where it can no longer apply. The radius has to be greater than 0 in any situation you can apply the formula.

For example, you can consider all the Earth surrounding you, and determine the gravitational effects of that; but the gravitational effect of something occupying the same location as you is undefined when using that formula. As Anders suggests in his answer; you can look at the limit as it approaches that point to figure out what you would expect to happen.

• Does the downvoter care to explain?
– JMac
Aug 28, 2018 at 20:19
• I must be the second down-voter. You might be trying to make a valid point, but your choice of wording makes your statements incorrect. It's not true that Newton's Law of gravity does not apply. What is true is that it has to be applied carefully. And when that is done, we find that the law can be used as $r\rightarrow 0$ . You might consider reworking your answer. Aug 28, 2018 at 20:41
• Using it when r approaches 0 isn’t the same as trying to apply it at r=0. Newton’s law of gravitation determines the force between two objects with mass; which cannot have the exact same location in Newtonian physics. My answer doesn’t say Newtonian gravity can’t determine an answer to this, just that trying to apply it to r=0 doesn’t work and wouldn’t represent a physical situation anyways.
– JMac
Aug 28, 2018 at 20:49