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.

 

[asteriatic]

Your thinking is correct, there is a key factor that you are missing in your calculations. While it is true that the gravitational force between two objects is directly proportional to their masses, there is another factor that comes into play - the distance between the two objects.

The gravitational acceleration that an object experiences is not solely determined by its own mass, but also by the mass of the object it is falling towards and the distance between them. This is described by the equation F = G(m1*m2)/r^2, where G is the universal gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

In the case of an object falling towards Earth, the distance between the object and the Earth's center is relatively constant, so the gravitational acceleration remains relatively constant as well. This is why, regardless of the mass of the falling object, the gravitational acceleration towards Earth is approximately 9.8 m/s^2.

On the other hand, if we were to consider the gravitational force between the falling object and the Earth, we would see that it does indeed change with the mass of the falling object. This is because the force is directly proportional to the masses of the two objects.

In summary, the gravitational acceleration experienced by a falling object is determined by the mass and distance between the object and the Earth, while the force between the two objects is determined by the masses of both objects. This is why the gravitational acceleration towards Earth remains constant, even as the mass of the falling object changes.