Why Doesn't the Earth Move When Objects Fall?

[Bashyboy]

The question I am looking at is: "When an object falls freely under the influence of gravity there is a net force mg exerted on it by the earth. Yet by Newton's third law the object exerts an equal opposite force on the earth. Why doesn't the Earth move?"

My actual problem isn't with the question, but with the solution, which I read after I solved it. To solve the problem, the gave a simple calculation to find the answer. The imagined a 1 kg object, and they approximated Earth's mass to be of the order of 10^25 kg.
They stated that $F_{_{Earth}} = F_{_{Object}}\rightarrow(10^{25} kg)a = 1 kg\cdot9.8m/s^2$ And solving for a would give $9.8\cdot10^{-25} m/s^2$

So, the gravitational acceleration the object provides would be that minute number. But say the object had a mass of 100 kg. Re-doing the calculations would yield $9.8\cdot10^{-23}m/s^2$ Why does this gravitational acceleration change? Shouldn't Earth's gravitational acceleration change then? I now my thinking is erroneous somehow, but I can't seem to figure out why. Does it maybe have to do with Earth's mass be constant? I'm not entirely sure.

 

[Doc Al]

They stated that $F_{_{Earth}} = F_{_{Object}}\rightarrow(10^{25} kg)a = 1 kg\cdot9.8m/s^2$ And solving for a would give $9.8\cdot10^{-25} m/s^2$

OK, that's the acceleration of the Earth due to the object's mass.

So, the gravitational acceleration the object provides would be that minute number. But say the object had a mass of 100 kg. Re-doing the calculations would yield $9.8\cdot10^{-23}m/s^2$ Why does this gravitational acceleration change?

The force on it (the Earth) increased by a factor of 100, yet its mass remained the same.

Shouldn't Earth's gravitational acceleration change then?

I assume you mean: Shouldn't the acceleration of the object change? No, since both force and mass increased by the same factor.

 

[Manan Shah]

There is a lot easier way to do the problem without equations, but just with common knowledge.

Newton's third law states that every object exerts an equal and opposite force on another.

The force that the ball exerts on the Earth is the same as the force that the Earth exerts on the ball, so why does the Earth not move?

The reason involves one simple formula: F = ma. This formula can be manipulated to read a = F/m, where a is the acceleration, F is the force, and m is the mass.

The ball's acceleration is (force) / (small mass) = HIGH ACCELERATION
The Earth's acceleration is (force) / (huge mass [5.9742 × 10^24 kg]) = SMALL ACCELERATION

Thus, the Earth DOES move up to meet the ball, but at such a small distance that we cannot see it.

Hope this helps!

Thanks,
Manan

 

[SammyS]

The question I am looking at is: "When an object falls freely under the influence of gravity there is a net force mg exerted on it by the earth. Yet by Newton's third law the object exerts an equal opposite force on the earth. Why doesn't the Earth move?"

My actual problem isn't with the question, but with the solution, which I read after I solved it. To solve the problem, the gave a simple calculation to find the answer. The imagined a 1 kg object, and they approximated Earth's mass to be of the order of 10^25 kg.
They stated that $F_{_{Earth}} = F_{_{Object}}\rightarrow(10^{25} kg)a = 1 kg\cdot9.8m/s^2$ And solving for a would give $9.8\cdot10^{-25} m/s^2$

So, the gravitational acceleration the object provides would be that minute number. But say the object had a mass of 100 kg. Re-doing the calculations would yield $9.8\cdot10^{-23}m/s^2$ Why does this gravitational acceleration change? Shouldn't Earth's gravitational acceleration change then? I now my thinking is erroneous somehow, but I can't seem to figure out why. Does it maybe have to do with Earth's mass be constant? I'm not entirely sure.

The Earth does accelerate towards the object, but with an extremely small acceleration.