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Join this group to post and comment. Kaustubh Katdare • Mar 1, 2007

Question On Gravitational Force

Few days ago, I thought of a question that tickled my common sense 😀 . It took me few minutes to actually refresh my basics about gravitational force, but finally I found the answer to this simple question: -

First, the homework:-
Gravitational force between any two masses is directly proportional to the product of the masses & inversely proportional to the square of the distance between them.

F = G (M1 * M2) / r^2

Let us consider that two masses of 1kg & 100kg are kept at a distance of 100 m from the earth. Obviously, as the distance (r) is same for both the masses, from the earth is same, we can neglect it for the time being.

According to formula, the force exerted on 100Kg mass will be 100 times more than the force exerted on 1kg mass. Therefore, it should come down to earth faster 😁 & reach the ground earlier than the 1kg mass, what say?

But we all know that both the masses will hit the ground at the same time. What's wrong in this argument?

Its easy! Use common sense & answer!

-The Big K- F=MA

does that suffice? 😎

well the description is that the accelaration will be equal as force is devided by mass and no drag by medium is present (masses considered to be point mass) Kaustubh Katdare • Mar 14, 2007
F=MA

does that suffice? 😎

well the description is that the accelaration will be equal as force is devided by mass and no drag by medium is present (masses considered to be point mass)
That, of course, is true. 😀

I'm looking forward to an answer which explains why the argument I made in the first post of this thread is incorrect 😁 . Of course I'm in no mood of proving the law incorrect .

Its simple! But can anyone put it in words?

-The Big K- th3 ied kid • Mar 16, 2007
well simplly while you try and derive with differential equations(or otherwise as well) for the time taken by the mass in the gravitational field of the planet the mass term gets cancelled ruining our logic of mass determinig time! xheavenlyx • Mar 20, 2007
Ok lets see. Did every try to solve this Q or just shoot anserws by reading it? 😛 😛

ok, lets calculate:

For mass 100kg : the answer is
3.987300564e12

For mass 1kg : the anser is
3.987300564e10

Hence we can say, for small masses we can safely assume the Force to be equal. But for masses greater then about 3/4th of earths we can take it in consideration (This is mine hypothesis...dont know for sure when we take smaller mass in calculations)

Good Q though Biggie 😀 Kaustubh Katdare • Mar 20, 2007
xheavenlyx
Ok lets see. Did every try to solve this Q or just shoot anserws by reading it? 😛 😛

ok, lets calculate:

For mass 100kg : the answer is
3.987300564e12

For mass 1kg : the anser is
3.987300564e10

Hence we can say, for small masses we can safely assume the Force to be equal. But for masses greater then about 3/4th of earths we can take it in consideration (This is mine hypothesis...dont know for sure when we take smaller mass in calculations)

Good Q though Biggie 😀
Think Simple! 😁

The hypothesis that you presented isn't true (or do we have Newton Jr. on CE? ).

Actually, the answer is very simple. I'll wait for some more time before posting the answer.

Btw, try asking the same question to your friends and see how they react 😉

All the best!

-The Big K- JimLFisher • Apr 3, 2007
The force on the 100 kg mass is 100 times as great. However, as the mass is 100 times more, the acceleration is the same. As velocity is produced as a product of acceleration and time, as long as they are both falling from the same height, they are thus falling for the same amount of time and thus both strike the earth at the same velocity. This, of course ignores the retarding effect of air resistance. Kaustubh Katdare • Apr 3, 2007
JimLFisher
The force on the 100 kg mass is 100 times as great. However, as the mass is 100 times more, the acceleration is the same. As velocity is produced as a product of acceleration and time, as long as they are both falling from the same height, they are thus falling for the same amount of time and thus both strike the earth at the same velocity. This, of course ignores the retarding effect of air resistance.
😁😁

That's exactly what I was looking for! You cracked it, Jim! -The Big K- technospartan • Feb 27, 2008
The_Big_K
😁😁

That's exactly what I was looking for! You cracked it, Jim! -The Big K-
that was good one dude. we must really focus on basic concepts