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High school physics
Course: High school physics > Unit 7
Lesson 8: Gravitational potential energy at large distancesGravitational potential energy at large distances
Learn how to calculate gravitational potential energy when we can no longer assume the gravitational field is uniform.
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I read somewhere that aboslute gravitational potential energy is negative because it is workdone against gravity and that energy of an object in bound state is negative so is it the reason or we just take it as negative for our own convenience?(11 votes)- confused: r^2 in newtonian definition is earth radius. but r @2:50is height or distance. So these are differnt things--how do they cancel out??(4 votes)
r is often approximated as earth's radius for situations happening close to the surface but it is really the distance between the two centres of mass(4 votes)
- Energy can't have a negative value. How come Ug= -(G*m1*m2)/r ?(2 votes)
- Hello there, Ryunah Kang,
It is a common misconception that energy cannot have a negative value. You see, when we refer to the object that Sal talked about at2:54we do so by refering to its to total gravitational energy with respect to the space—time continuum.
Ideally, Space-time when there exists no mass around a space-time continuum, there should be no deformation. If we consider mapping this plane, we can make its displacement (height or in the case of Sal's example at2:54, r) 0. So at this point r=0.
But when mass is introduced on to the continuum, it bends the fabric of space time. This allows for a gravity well to be produced. When an object experiences the force of this gravity well, it moves in its direction: downwards (with respect to the point where r=0) hence the Ug value is negative with respect to space—time.(5 votes)
the value of g=9.8 m/s2, does it include any air resistance?(1 vote)
The minuscule mass of the air on Earth could add to the overall mass of the planet when considering gravitational acceleration at large distances, but no, air resistance does not play a factor with gravity in this case.(4 votes)
Does variation of g happen with respect to altitude ?(1 vote)
Yes, but the change in the gravitational acceleration is so miniscule and very rarely do you need to be more precise than 3 significant figures with this value. It'll take heights bigger than the peak of Mt. Everest to change the value of g even by 1e-2m/s^2.(3 votes)
g=GM/R^2
here g is inversely proportional to the R square. That means if radius will increase then acceleration due to gravity will decrease. But how is it possible?(1 vote)
The farther you get from the Earth, the less force you experience and therefore the less acceleration you experience. Hope this helps!(3 votes)
- @4:10Is mgy considered a negative force?(1 vote)
- First off, mgy is not a force. It is the energy of gravitational potential on the object. This value is not negative as it is in reference to the height above the planet. Usually, the value of Ug is negative because the height of an object suspended in space-time is always negative due to the bending of space-time due to mass(2 votes)
- So is the 'y'- which is the height, irrelevant then?(1 vote)
Well, the variables were kinda confusing but essentially the r=y. In the newtons law of gravitation equation the variable r it is normally used for radius of the Earth but now because we are dealing with these long distances, it is important to note that the r is actually the distance between the 2 objects. Hope, it helped:)(1 vote)
Is there a proof for this formula?(1 vote)
How can you find out what energy they have and how to find their value? I'm just really confused.(1 vote)
2:50Video transcript
- [Tutor] Let's do a little bit of a review of potential energy and especially gravitational
potential energy, because in this video, we're going to get a
little bit more precise. So let's say that I have an object here, it has a mass of M and I were to change its position
in the vertical direction, we're assuming that we are on Earth, where the gravitational field is G and we have a change in the
vertical direction of delta y, so this is its new position,
the object of mass, M, so what is its change in
gravitational potential energy? Well, we have seen this before, our change in potential
energy, due to gravity is going to be the mass of our object times the gravitational field times our change in position
in the vertical direction. Now this is all good, but
there's a couple of things, that we can get a little bit more precise, that we can think about in this video, one is we assumed a
constant gravitational field and that's fine if we're
near the surface of a planet and we also are just doing
everything on a relative basis, but is there a way to come
up with an absolute number on what is the gravitational
potential energy? Well, to think about that, let's just go back to
Newton's Law of Gravity, Newton's Law of Gravity tells
us that if I have two objects, so this object has mass, M1, this object has mass, M2 and the distance between
their center of masses is r, that the magnitude of the force of gravity and if the force of gravity's acting in this direction on that object, it would have an equal
and opposite direction on that object right over there, so the magnitude of the force of gravity is going to be equal to
the gravitational constant times the product of
the two objects' masses divided by the radius squared or another way to think about it, if we wanna know the
gravitational field created by M2, well then you would just
divide both sides by M1, so if you divide both sides by M1, you get that the gravitational field actually isn't constant, it is equal to the gravitational constant times, these cancel out, M2, the mass of the object generating
the gravitational field divided by r squared and so this can actually help us come up with an absolute formula for gravitational potential energy. We can say that gravitational
potential energy and this is typically denoted
with a capital G over here is equal to, let's say
the mass of the object and let's just call that M1, the mass of the object times the gravitational field, so that's G times the mass of
the object creating the field divided by r squared times the height that this is above the center of mass of the object generating the gravitational field, well, what's that? Well, if this was Earth,
you would view r as, well, how far am I above
the center of mass of Earth? And so you would multiply that times r, times r and so this r and one of
the rs in the r squared will cancel out, so that will just go away and then this will just become an r and now one thing to think about is what should the potential energy be as r goes to infinity? Well, what is the gravitational
field as we go to infinity, as r gets really, really, really large? Well, as r gets really,
really, really large, the gravitational field goes to zero, but at the same time,
the further you get away, the higher the potential
energy you should have and so the way that
we've thought about this is we say, hey look, let's
just put a negative out front, you could just put it right over here and then it all works out, the gravitational potential energies are always gonna be negative, but they get less negative,
the further out you get to and at infinity, it gets to zero. So let me rewrite this. The potential energy, due to gravity can be written as the negative
of the gravitational constant times the mass of the first object time the mass of the
second object divided by r and you can see this looks an awful lot like Newton's Law of Gravity, what the difference is, we have a negative in front
of the gravitational constant, instead of dividing by r
squared, we only divide by r, but this is really, really,
really valuable now, because now we have an absolute sense of what gravitational potential energy is and it meets all the properties we need, it would still meet this idea
that the further you get away, you will have an increase in
gravitational potential energy, because it's going to
become less negative, but at the same time
as you go to infinity, as r gets infinitely far, your potential energy is
going to approach zero. So I will leave you there, in future videos, we will apply this.