# Can someone please explain this statement regarding Newton's 3rd law?

This is a statement from my textbook
"When an object falls down, it falls inwards towards the centre of mass of the Earth pulled by it's gravity. As said in Newton's 3rd law of motion the falling object will also exert an equal and opposite force on Earth; which then accelerates upwards towards the object."

What the heck does that mean?

If a satellite falls down on earth and the gravity on it is 9.8 then is it also exerting a force of 9.8 on earth? How is that possible?

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Ibix
You are mixing up force and acceleration; otherwise you are doing fine.

The Earth exerts a force of 9.8N on a 1kg mass, causing it to accelerate towards the Earth at 9.8ms-2. The mass exerts a force of 9.8N on the Earth, causing it to accelerate towards the mass at 9.8/6x1024=1.6x10-24ms-2. The forces are the same, but the visible results are very different.

If you're still boggling at that, ask yourself a question: what's the other option? That the Earth doesn't move? If so, why does it move under the gravity of the Sun? You could also look up the word barycentre.

Another way to look at it is that it's all just atoms pulling on each other. Some of the atoms happen to be bound together in large balls, and some in small balls, but it's all just atoms.

So F=ma and the force exerted depends on mass since the earth has a much larger mass we barely feel it but the force exerted on the object by it is felt because that object has a much smaller mass?

But then what is with equal and opposite forces?

HallsofIvy
Exactly what it says- the force of gravity between to masses is $-GMm/r^2$. Each object pulls on the other with exacty the same force. If the earth pulls on an object with mass m, then, because "F= ma" so a= F/m, its acceleration is $-GM/r^2[/tex]. The mass "m" has canceled. And the object pulls on the earth with exactly the same force. The difference is that now, to find the earth's acceleration we divide by earth's mass which is much much larger- the earth's acceleration toward the object is [itex]-Gm/r^2$ which is too small to be measured.