0
$\begingroup$

I am doing Newtonian Physics. Correct me if I am wrong: So two objects always exert equal forces on each other(Newtons Third). So for example the same gravitational force that the earth exerts on a tennis ball, a tennis ball exerts on the earth. We also know that the acceleration of gravity is irrespective of mass and on earth, it is 9.8 ms. Shouldn't the earth also accelerate to objects at 9.8 ms?

So my question is: Isn't the Earth constantly accelerating to all objects on it at the fast rate of 9.8ms? Should we notice the earth moving around like that?

$\endgroup$
2
  • 1
    $\begingroup$ Force and acceleration are two different things. $\endgroup$ – ACuriousMind Nov 15 '15 at 14:04
  • 1
    $\begingroup$ the forces are equal, but the accelerations differ because a=F/m, so each object accelerates at a different rate. $\endgroup$ – user83548 Nov 15 '15 at 14:23
2
$\begingroup$

No. the earth do not accelerate with 9.8 meter per sec. squared. The acceleration is quite negligible. And as the objects on the whole earth is distributed almost uniformly the net acceleration due to the objects is zero. That's why we do not notice the earth moving around like that.

$MATHEMATICALLY$

we know that

$| F_E | = |F_o|$

where $F_E$ is force exerted by earth on the object. And $F_o$ is force exerted by object on the earth.

Now, therefore

$m_E a_E = m_o a_o$

$a_o = m_E a_E/m_o$

As $m_E$ >>>> $m_o$ ...So $a_o$ >>>> $a_E$

So we can say that $a_E$ is almost zero due to a single object.

$\endgroup$
0
$\begingroup$

The flaw in the logic of the o/p's question is more easily stated thus: the Earth has a given mass, and the tennis ball has a given mass, but each mass has a different value.

The Earth has planetary mass, according to which an object nearby in free fall will accelerate toward it at 30 ft per second per second. But the tennis ball has a mass of about 6 oz, so has a negligible acceleration upon an object nearby.

You cannot simply say that the acceleration of 30 ft/sec applies equally in both directions: simple logic will tell you that it applies only in one direction, since it is solely caused by the mass of the more massive body.

If the Earth was removed from the equation, no one would expect the mass of the tennis ball to exert any (non-trivial) acceleration upon any nearby object. So one could scarcely expect the Earth to be accelerated by the ball's (negligible) mass.

Furthermore, the logic offered by the o/p leaves out of account entirely the very obvious fact that inertia will hold the Earth in place. Unless the mass of the other object has a significant fraction of the Earth's mass, the inertia associated with the Earth's mass will prevent it from being accelerated in any direction by the presence of the other object.

$\endgroup$
0
$\begingroup$

No. Newton's second law tells us that the acceleration profile obtained by application of a force $F$ to an object should be given by

$$a = \frac{F}{m}$$

where $m$ is the object's mass. This means it is inversely proportional to mass. Moreover, it is Newton's third law which tells us that there must be something opposite happening to the Earth from the ball, but it is not that the ball produces an equal and opposite acceleration, but rather an equal and opposite force. If the ball has a mass of, say, 100 g, or 0.1 kg, but the Earth has a mass of 6000 Yg, or $6 \times 10^{24}$ kg, then that means that, thanks to the inverse proportionality, while the ball will accelerate at 9.8 $\mathrm{m/s^2}$, the Earth will accelerate about $6 \times 10^{25}$ times less strongly. That equates to about 0.16 $\mathrm{ym/s^2}$, or yoctometres per second squared. Equivalently, that is $0.16\ \mathrm{(m/s)/Ys}$, or 0.16 metres per second per yottasecond, meaning it will take an entire yottasecond - about two million times the age of the Universe - for the Earth to be accelerated by this force to even 0.16 m/s of speed. Of course, the ball has already contacted the Earth's surface long, long before that point and so the mutual acceleration of the two has ceased.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.