In the Laws of Gravity by Galileo, we say that every
object take equal time to fall if there's no air resistance.
We always ignore that **The Earth will also have
acceleration due to that object also !**
So, if we don't ignore that concept, **Are Galileo's Laws
of Gravity correct?** Think once again, can you explain
your logic behind your opinion?

Traditionally, \[g=\frac{\text{GM}}{r^2} \] Where, M=Mass of earth r=Distance between the object and the center of the earth

But in reality, earth will also have the acceleration due to the object of mass **m**. In that case, resultant acceleration,
\[g=g_o+g_e\]
Here, \(g_e\)= Acceleration due to earth (for the object)
And, \(g_o\)= Acceleration due to object (for the earth!)
So,
\[g_o=\frac{\text{Gm}}{r^2}\]
\[g_e=\frac{\text{GM}}{r^2}\]

Then, assume the earth is **still**. So,
**Relative (Actual) Acceleration of the object**
\[g=g_o+g_e\]
\[=\frac{\text{G}}{r^2}\text{(M+m)}\]

The equation says that *Heavy objects have greater relative acceleration compared to lighter objects*.
So, heavy objects falls **earlier** !

So, in reality, in actuality, **Galileo's Laws aren't perfect!**

But most of the time we ignore acceleration of the earth (Acceleration due to object) , which makes calculations simpler!

## Comments

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TopNewestYou cannot add up those accelerations. By Newton's Third Law, they are equal amd opposite, which is perfectly okay because we're seeing one of them from the earth's coordinates and the other from the object's reference frame. When you mean to add them up, what reference frame are you using?

If this is unclear with A, let's look at velocities. If A is moving towards B with speed 5, it should appear that B is also moving towards A with speed 5. So, would that mean A is approaching B with velocity 10? – Agnishom Chattopadhyay · 1 year, 10 months ago

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@Arifur Rahman This equation is very hard to understand:

\[g = g_o+g_e\]

What object is supposed to have the acceleration \(g_o+g_e\) in your situation? – Josh Silverman Staff · 1 year, 10 months ago

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still, then the object is coming towards the earth with acceleration,g. Or, if you prefer so, assume the objectisn't moving, then the earth is reaching the object with acceleration,g.Its just to make hideous calculations simplest.

So, if you don't like to use any of the above techniques, do your own equations. But whatever you do, you can't neglect

Acceleration due to objectthis time.Actually, its much more conceptual, rather than being analytical. Make your brain thinking and everything will be crystal clear. – Arifur Rahman · 1 year, 10 months ago

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– Josh Silverman Staff · 1 year, 10 months ago

Oh, now I see what you are saying. If we hold the Earth, or the object still, then the acceleration of one of them relative to the other will happen at \(g\). How massive does the object have to be before we can notice this effect?Log in to reply

– Arifur Rahman · 1 year, 10 months ago

I think Josh Silverman, who's a problemsetter of the "IPhOO 2014" , knows the answer, without anyone's help.Log in to reply

– Agnishom Chattopadhyay · 1 year, 10 months ago

I still doubt that. Where did the g_0 come from in the first place, then?Log in to reply

Acceleration due to the object, as its already mentioned. And this is what youneglectin your Galilean Equations. – Arifur Rahman · 1 year, 10 months agoLog in to reply

– Josh Silverman Staff · 1 year, 10 months ago

How massive would the object need to be to change its measured acceleration by say 1%?Log in to reply

Your question is a little ambiguous - 1% change in g - increase or decrease? Well, obviously its

increaseThen \(g_2=g_1(1+.01)=1.01g_1\) In general, \(g_1=\frac{\text{GM}}{r^2}\) Using proportionality concept, \[\frac{g_2}{g_1}=\frac{\text{M}+m}{\text{M}}\] \[\rightarrow 1.01\text{M}=\text{M+m}\] \[\rightarrow m=0.01\text{M} \] You can take Mass of the earth, \(\text{M}=6\times 10^{24} kg\) Then, \(m=6\times 10^{22} kg\)Notice that, its quite a

Hugemass. Whichnecessarilymeans we canneglectit.But

Neglecting something doesn't mean the thing doesn't EXISTS!. – Arifur Rahman · 1 year, 10 months agoLog in to reply

I agree with Arifur's idea..he wants to say that both earth and the object attract each other..the object's acceleration depends on the earth's mass and the earth's acceleration on the object's mass.when we look from the earth at the object falling, what we are observing is its relative acceleration w.r.t. the earth frame which in turn depends on both by the relationship that Arifur has mentioned in his note. – Kuldeep Guha Mazumder · 1 year, 6 months ago

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– Arifur Rahman · 1 year, 6 months ago

Exactly. I wish we could have Galileo alive to correct him!Log in to reply

– Kuldeep Guha Mazumder · 1 year, 6 months ago

Galileo made only some experimental observations without knowing what law works behind.. and the error due to the earth's acceleration is so negligible that it can be hardly measured by experiments..further when someone is dropping a body at one location of the earth's surface..there may be bodies which are being dropped (or falling) at various different location..which may even nullify the first body's effect on the earth..so experiments can hardly detect it.Log in to reply

gby only 1%. Most probably this isn't measurable anyway. And when there are thousands of objects being dropped every moment, nullification is so common. We can't even witness such difference. – Arifur Rahman · 1 year, 6 months agoLog in to reply

– Kuldeep Guha Mazumder · 1 year, 6 months ago

Yup! That's what I am saying.Log in to reply

We were taught in the school that irrespective of the weight, when two objects of different weight are thrown at the same time from the same height in an ideal resistance-less atmosphere both would reach the earth at one and the same time as per Galileo's Law.I had my own doubt though I could not disprove this Galileo Law and felt that heavier object would reach the earth slightly earlier than the lighter one, even though the difference in time could be too small to be taken into account. Arifur Rahman's proof as above seems to vindicate what I had doubted. One formula in Physics says F=ma, where F is the force and m and a are mass and acceleration respectively of the objects. That means Force (F) generated is a function of mass (m) and acceleration (a). Heavier the mass or more the acceleration or both more would be the Force exerted. Obviously though the acceleration remains the same and the mass varies or if mass remains the same and acceleration varies, the heavier object and the object with higher acceleration would be acted upon by more force than the lighter one/or with less acceleration and hence the heavier object with more acceleration would move a bit faster and reach the earth earlier than the lighter one howsoever smaaaaaal the difference in the force is. Further, every object in the Universe attracts the other objects even if the force of attraction is too small to be felt. As such, the falling objects attract the earth and so also the the earth the falling objects. This force also affects the motion of the falling objects. The Heavier will be attracted by more force and the lighter one with less force. This also makes their motion slightly different one from each other. In the end, I am inclined to agree with Rahman when he says when we ignore something because of its negligible value does not mean it does not exist. Hence we may say that the Galileo Law, though almost nearest to the actual finding, it has some minor difference which we have been ignoring. I invite Arifur Rahman to comment on what I have stated and add his views to this. – Venkatesh Patil · 1 year, 9 months ago

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thoughtfulexplanation. Its nearly correct, but not totally accurate. Yes, we don't need to consider earth's acceleration. But if you wantpinpoint correctanswer, you will have to do a little bit more.Thank you, Venkatesh Patil for your nice writeup which made this note precise.

BTW, If you're not in the Brilliant Lounge , join today. We will be happy to have you onboard. – Arifur Rahman · 1 year, 9 months ago

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You're looking and searching here and there for a Reference Frame, right?

Well, make the universe your reference frame (though nothing in this holy Universe is stationery!) . You'd like some more calculations. You might've solved

Train Collisiontype problems. Both trainsdeceleratetoStop a collisionfrom a distance(linear, of course)d. What Reference Frame you use there? - The surroundings , to be more preciseEarth is the reference framethere.So, if the distance between

Center of the earthand theObjectisrYou'll make, \[r=\frac{1}{2}g_et^2+\frac{1}{2}g_ot^2\] \[r=\frac{1}{2}(g_e+g_o)t^2\] I think its excess to explain how we came up here. And \((g_e+g_o)\) is that so calledRelative Accelerationand you can find out millions ofReference FramesaroundEarth- Object AttractionIts a bit surprising that

Agnishomis asking this type of questions. Because your Physics Level isrelativelyhigher that average explorers on Brilliant. – Arifur Rahman · 1 year, 10 months agoLog in to reply

– Agnishom Chattopadhyay · 1 year, 10 months ago

If the earth is the frame of reference, are we not supposed to see it still?Log in to reply

The surroundingsaround earth and the object is that reference frame. If earth was the RF, then what I'm talking about! – Arifur Rahman · 1 year, 10 months agoLog in to reply

@Agnishom Chattopadhyay pls check out this note. – Ashley Shamidha · 1 year, 10 months ago

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– Agnishom Chattopadhyay · 1 year, 10 months ago

Thanks for the mention. I find something fishy here.Log in to reply

You could've found a better word for the comment. Its like, you're forcing the reality like Catholic Churches. – Arifur Rahman · 1 year, 10 months ago

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– Agnishom Chattopadhyay · 1 year, 10 months ago

I'm sorry you're offended by that word. I meant that your idea looks a bit wrong.Log in to reply

– Arifur Rahman · 1 year, 10 months ago

Thank you. That's called frankly speaking, being friendly over the internet. You're appreciating.Log in to reply

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– Arifur Rahman · 1 year, 10 months ago

Its really a bit astonishing.Log in to reply