Terminal velocity

[axp107]

1) Can someone explain what I need to know about terminal velocity for the MCAT. What exactly is it.. everytime I see the word terminal velocity in a passage, I don't know what to do...

Seems like it should be easy points. Any tips?

2) What is escape velocity, then.

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[sehnsucht]

1) Can someone explain what I need to know about terminal velocity for the MCAT. What exactly is it.. everytime I see the word terminal velocity in a passage, I don't know what to do...

Seems like it should be easy points. Any tips?

2) What is escape velocity, then.

terminal velocity would be when a falling object reaches equilibrium in motion, ie. the acceleration due to g equals the normal force for a given object. the net force for an object experiencing terminal velocity is zero.

escape velocity is the velocity required for an object to escape the pull of gravity from, say, the earth. in a passage on the MCAT, terminal velocity would be mentioned in a problem dealing with a rocket or something of the sort. all objects on earth are subject to the same escape velocity regardless of their mass - what differs is the force required to accelerate an object to a specific velocity sufficient enough to reach the escape velocity. escape velocity is also subject to the mass of the planet. to compare, the escape velocity of earth is significantly smaller than Jupiter's.

 

[vicinihil]

terminal velocity would be when a falling object reaches equilibrium in motion, ie. the acceleration due to g equals the normal force for a given object. the net force for an object experiencing terminal velocity is zero.

escape velocity is the velocity required for an object to escape the pull of gravity from, say, the earth. in a passage on the MCAT, terminal velocity would be mentioned in a problem dealing with a rocket or something of the sort. all objects on earth are subject to the same escape velocity regardless of their mass - what differs is the force required to accelerate an object to a specific velocity sufficient enough to reach the escape velocity. escape velocity is also subject to the mass of the planet. to compare, the escape velocity of earth is significantly smaller than Jupiter's.

you mean escape velocity right? not terminal velocity

Is there some sort of equation for this? Terminal velocity: a = Fn = mg? So if I weighed 100g, my max acceleration is 100 * g or 98 m/s^2...so if I were in free fall for 4 seconds...my terminal velocity is then...4 * 98 or about 400m/s? Am I on the right track?

 

[inside_edition]

1) Can someone explain what I need to know about terminal velocity for the MCAT. What exactly is it.. everytime I see the word terminal velocity in a passage, I don't know what to do...

Seems like it should be easy points. Any tips?

2) What is escape velocity, then.

terminal velocity is when air resistance equals gravity. therefore your net acceleration is zero and your speed does not change. compare a paper ball to a basketball: you simply can't get a paper ball to go fast but you can with a basketball.

escape velocity=how fast you must be going in order to get on a planet's orbit. for earth it is 5 miles per second.

 

[axp107]

1) In a given physics problem, how would you calculate terminal velocity?

Set mg = to F of air resistance... then what?


2) How would you find escape velocity using the GMm/r^2 formula?

 

[BrokenGlass]

1) In a given physics problem, how would you calculate terminal velocity?

Set mg = to F of air resistance... then what?


2) How would you find escape velocity using the GMm/r^2 formula?

Escape velocity: GMm/r^2 = mv^2/r; solve for v to get the escape velocity.

I have also seen cases where escape velocity for the Earth is defined as the velocity needed to reach an infinite distance from the Earth. U = -GMm/r. So the maximim graviational potentail energy is zero (when you are infinitely far apart from the Earth). So the higher your initial veloctiy, the higher is your kinetic energy (KE = 1/2mv^2). When you are "escaping" from the Earth, you are just converting KE to U, where U stands for (gravitational) potential energy.

Terminal velocity: an object reaches terminal velocity when the force on this object due to air resistance is exactly equal and opposite to the object's weight. So the object is in dynamic equilibirum, i.e. it's moving at constant (aka terminal) velocity.

HTH

 

[DrDuber]

1) In a given physics problem, how would you calculate terminal velocity?

Set mg = to F of air resistance... then what?

Force due to air resistance= (Cross sectional area of object)(Constant of fluid found empirically)(Velocity of object)

set this equal to mg and solve for the velocity. this is usually the form found but in some cases the force of resistance scales as velocity^2

 

[sehnsucht]

you mean escape velocity right? not terminal velocity

Is there some sort of equation for this? Terminal velocity: a = Fn = mg? So if I weighed 100g, my max acceleration is 100 * g or 98 m/s^2...so if I were in free fall for 4 seconds...my terminal velocity is then...4 * 98 or about 400m/s? Am I on the right track?

YES! thanks for pointing out the type-o. In the MCAT, terminal velocity would be contextualized in an equilibrium problem where questions would ask what net forces are acting on an object in terminal velocity. The passage will have other useless details embedded in it but all you need to do is pay attention to the words constant velocity because it would indicate that there are no net forces acting on the object since the acceleration due to g equals the normal at terminal velocity.

Escape velocity, on the other hand, would be addressed in a question that deals with rocketry or something where you would need to calculate how much force would be required to accelerate an object so it overcomes the gravitational force of the earth to go into space. The equation for this is derived from the Fg = GMm/r^2 equation where Fg (force due to gravity) is equal to the kinetic force equation 1/2mv^2. The equation expresses that the velocity required of a rocket to attain escape velocity is equal to the force g that exists between the rocket and the earth.

 

[artaxerxes]

For escape velocity,

KE=Ug infinity-Ug at surface=0-Ug at surface, since Ug surface is negative, this value is positive as expected.

KE=mgr r=radius of earth or whatever object
V=(2gr)^1/2

(2gr)^1/2=(2(GM/r^2)(r))^1/2=(2GM/r)^1/2

when mass of escaping object is negliable

 

[axp107]

I dont understand what the above poster did.

 

[artaxerxes]

you know that potential energy=mgh

for escape V, h=r=radius of earth

KE=1/2 mv^2

to escape, KE has to equal potential energy or be greater

so 1/2 mV^2= mgh, divide by m, divide by 1/2, square root and you get

V=(2gh)^1/2

the second half is from setting g=F=GMm/r^2 for the equation mgr, when m is negliable, you get V=(2GM/r)^1/2

 

[axp107]

What if you set GMM/r^2 equal to mv^2/r?

Then we'd get velocity = (GM/r)^1/2

 

[BrokenGlass]

What if you set GMM/r^2 equal to mv^2/r?

Then we'd get velocity = (GM/r)^1/2

v = (GM/r)^1/2 describes the velocity needed to orbit the Earth at a radius r, where r is measured from the center of the Earth. "Escaping" in this context means "becoming a satellite."

Another way I have seen "escaping" defined is getting infinitely far from the Earth, in which case the gravitational pull of the Earth on the escaped object is zero because F = GMm/r^2, and as r goes to infinity, F goes to zero. U = -GMm/r, where U is gravitational potential energy. The largest value of U is zero and this value is attained when a body is infinitely far away from the Earth. In order to get infinitely far away from the earth, an object needs to have enough kinetic energy (and therefore enough velocity) and as it is getting further away from the earth, its KE is being converted to U.

I don't think it's correct to use mg in escape velocity problems because g is just an approximation for cases when a body is close to the surface of the Earth.

Anyhow, perhaps someone with a good understanding of this can post more (and/or correct me if I am wrong)

 

[artaxerxes]

v = (GM/r)^1/2 describes the velocity needed to orbit the Earth at a radius r, where r is measured from the center of the Earth. "Escaping" in this context means "becoming a satellite."

Another way I have seen "escaping" defined is getting infinitely far from the Earth, in which case the gravitational pull of the Earth on the escaped object is zero because F = GMm/r^2, and as r goes to infinity, F goes to zero. U = -GMm/r, where U is gravitational potential energy. The largest value of U is zero and this value is attained when a body is infinitely far away from the Earth. In order to get infinitely far away from the earth, an object needs to have enough kinetic energy (and therefore enough velocity) and as it is getting further away from the earth, its KE is being converted to U.

I don't think it's correct to use mg in escape velocity problems because g is just an approximation for cases when a body is close to the surface of the Earth.

Anyhow, perhaps someone with a good understanding of this can post more (and/or correct me if I am wrong)

g is a good approximation at most positions near earth's surface. yes the gravitational force would get less as you increase your altitude, but I think for the vast majority of MCAT questions this increase is very negliable. Either case, the second formula, V=(2GM/r)^1/2 holds for all values of r I believe....for example, at R=infinity, V=0, which is what you would expect.

You can go to wikipedia's entry on escape velocity to see how its derived (calculus alert).

What if you set GMm/r^2 equal to mv^2/r?

Then we'd get velocity = (GM/r)^1/2

yes you would, but thats the velocity needed for orbit.
what you are doing is Fgravity = F centripetal, GMm/r^2=m(V^2/r)<--that is centripetal acceleration