Speed of object when gravitational force acts between two object

[Hardik Batra]

Suppose I have two object. One is 1 kg and other one is 2 kg. And separation between two object is 10 m . By gravitational force both objects attracts towards each other with some speed. But i want to know with what velocity they are moving towards each other.

Or is there any equation to find velocity of object?

Thanks.

 

[sophiecentaur]

1. You have a force (the same magnitude on each) so you have the acceleration (a = F/m)
2. v = u +at, which will tell you the velocity at time t (initial velocity, u will be zero) (Look up "SUVAT" equations - they are all over the web, at various levels if you want more)
3. Momentum is conserved, so the ratio of the velocities will be the inverse ratio of the masses.

 

[HallsofIvy]

Suppose I have two object. One is 1 kg and other one is 2 kg. And separation between two object is 10 m . By gravitational force both objects attracts towards each other with some speed.

No. Gravitational force attracts both objects toward the other with some acceleration, though NOT the same acceleration, which increases as the distance between them decreases.

But i want to know with what velocity they are moving towards each other.

There is no such velocity. You can say that the force is given by [itex]F= GmM/r^2[/quote] or $G(1)(2)/r^2= 2G/r^2$. That is, if we take our coordinate system to have 0 at the 1 kg object, so it is stationary in that coordinate system, then the other is attracted to it with acceleration $a= F/2= G/r^2$. Since a= dv/dt, dv= adt= a(dt/dr)dr= (a/v)dr so that $vdv= adr= Gr^{-2}dr$. Integrating, $(1/2)v^2= -G/r+ C$.<br /> <br /> Since v= 0 when r= 10, we have (1/2)(0)= 0= -G/10+ C and C= G/10. So $(1/2)v^2= G/10- G/r$ and $v= \sqrt{2G(1/10- 1/r)}$.<br /> <br /> That gives v as a function of r. To get it as a function of t, we would have to integrate again in order to get r as a function of t.[/itex]

 

[Hardik Batra]

Since v= 0 when r= 10, we have (1/2)(0)= 0= -G/10+ C and C= G/10. So $(1/2)v^2= G/10- G/r$ and $v= \sqrt{2G(1/10- 1/r)}$.

Suppose both particle comes at 5m distance then what are the velocity of two particle.

I think this..
$v= \sqrt{2G(1/10- 1/r)}$

put r=5 then

$v= \sqrt{2G(1/10- 1/5)}$

and the answer is

$v= \sqrt{2G * (-0.1)}$

then what next ?

 

[D H]

That is, if we take our coordinate system to have 0 at the 1 kg object, so it is stationary in that coordinate system, then the other is attracted to it with acceleration $a= F/2= G/r^2$.

That's not correct, Halls. You have chosen to use a non-inertial frame. Each of the two objects is accelerating toward the other. The relative acceleration of two free objects toward one another is $\frac{G(m_1+m_2)} {r^2}$.

Suppose both particle comes at 5m distance then what are the velocity of two particle.

I think this..
$v= \sqrt{2G(1/10- 1/r)}$

put r=5 then

$v= \sqrt{2G(1/10- 1/5)}$

and the answer is

$v= \sqrt{2G * (-0.1)}$

then what next ?

I think you ought to start over. An imaginary velocity doesn't make a bit of sense.

It looks like you are assuming the two objects are initially at rest with respect to one another. That's fine, but you should have spelled that assumption out.

Hint: What does conservation of energy have to say about this scenario?

 

[Doc Al]

Suppose both particle comes at 5m distance then what are the velocity of two particle.

I think this..
$v= \sqrt{2G(1/10- 1/r)}$

put r=5 then

$v= \sqrt{2G(1/10- 1/5)}$

and the answer is

$v= \sqrt{2G * (-0.1)}$

then what next ?

Careful. That will give you an imaginary velocity, so you know it's not quite right.

I suggest you start over and stick to an inertial reference frame. Apply conservation of energy: When the masses approach each other, the decrease in gravitational PE must equal the increase of total kinetic energy. (Since both masses move, you must include the KE of each.)

Combine that with conservation of momentum and you'll be able to solve for both velocities.

Edit: D H beat me to it!

 

[Hardik Batra]

Other Question is :

Momentum and kinetic energy are conserved when gravitational force acts between two objects ?

 

[MuIotaTau]

Other Question is :

Momentum and kinetic energy are conserved when gravitational force acts between two objects ?

Momentum is always conserved when no external forces are acting on the system. So as long as your system is the two objects, and no other objects are interacting with them, then yes, momentum is conserved.

Kinetic energy is not, in general, conserved, but the total energy is.

 

[Hardik Batra]

Momentum is always conserved when no external forces are acting on the system. So as long as your system is the two objects, and no other objects are interacting with them, then yes, momentum is conserved.

Kinetic energy is not, in general, conserved, but the total energy is.

Here, Gravity is acted upon two object that is external force, then how momentum is conserved.?

 

[MuIotaTau]

Here, Gravity is acted upon two object that is external force, then how momentum is conserved.?

So the system we're considering is the two objects, right? In this case, the gravity between these two objects is an internal force because it only involves objects in our system.

Now imagine a different situation: we still have our two objects, and they're in our system, but there's also a huge, third object outside of our system. In this case, the third object, which is outside of our system, exerts gravity on the two objects in our system, and this is an external force.

Does that distinction make sense?

 

[sophiecentaur]

Here, Gravity is acted upon two object that is external force, then how momentum is conserved?

If only the two objects are involved (no external forces) then total Momentum of the two objects is conserved. Before and until / if they collide the total energy (Kinetic + Gravitational Potential) is conserved.
If you want to introduce a third object then you need to sum the Momentum and Energy for all three objects. Total Momentum of the whole system is conserved and so is total mechanical Energy, until a collision occurs. The sums are just harder to do.

Afaiaa, there are no experiments which have shown a failure of Conservation of Momentum and, if you include the Mass / Energy equivalence, Energy is also conserved. But you have to include all forms of Momentum and Energy in the total. Any photons which may be emitted, carry away their own momentum, which can be relevant in the dynamics some reactions.