So when you have 2 objects with mass, they act on each other with gravitation force, in opposite directions. As you have forces acting in opposite directions, there is a point where the forces balance out, or the neutral point. This would be because if you imagine a line from object A to B, representing how the forces act. Starting at A, Force A will have the strongest force, but as you travel to B, the Force B will become stronger and A weaker as the distance r changes, to the resultant force cancels out more until it is 0. Then, when you carry on to B, Force B is the dominant force and Force A gradually cancels Force B out less until you get to point B.
A possible question to be asked is where this neutral point. If the masses were the same, then the point would be bang in the middle, but it usually isn't that simple. At the middle point g is 0 for both, so if you say r is the total distance between A and B, M is mass of A and m for B, and x is the distance from A to the neutral point, then GM/x²=Gm/(r-x)², which rearranges to M/m=x²/(r-x)², so given M, m and r you can find x.
Gravitational Potential
As you may notice, this term looks similar to Gravitational Potential Energy, which it is heavily related to, but not the same thing. As you may remember, the equation for GPE is Δ𝐸ₚ = 𝑚gΔℎ. Eₚ is the energy an object possesses because of it's height above the ground, though as we know more specifically now, the distance away from another object. The equation uses change in Eₚ as this is more practical for mechanics, but in this topic we can work out the actual 𝐸ₚ value of an object. At infinity distance away, 𝐸ₚ is 0, and as the object gets nearer, 𝐸ₚ gets more negative. This means that the 𝐸ₚ of an object in a field is actually negative, which does seem weird at first. However, remember that gravity is attractive, so always tries to pull objects to the centre. This means to lift an object up, against gravity, it takes work to push it away, so it gains energy (an its 𝐸ₚ gets closer to 0). When an object is falling, gravity is doing work, so the 𝐸ₚ gets smaller and more -ve.
You can also see it is negative by integrating Newton's law of gravitation. When you integrate Force with respect to distance, you get the energy, as W=Fd (∫F dr =E). When you integrate F=(GMm)/r², you get 𝐸ₚ=-(GMm)/r, so 𝐸ₚ is -ve or 0 (as r can't be <0). Also in this equation you can see why 𝐸ₚ is 0 at infinity, as doing r=-(GMm)/0 would be ∞
Gravitational Potential, V, of a point in a Gravitational field is defined as the work done per unit mass needed to move a smaller object from infinity to that point. The equation for this one is V=-GM/r. So at ∞ distance away, V=0, and as the object gets closer to the centre, V gets more -ve.
A little tangent (not in the mathematical sense), but this equation is the integral of gravitational field strength, g=GM/r², so ∫g dr =V. This is interesting as this is very similar to ∫F dr =E, it is just that it is times-ed my mass, so basically ∫mg dr =mV. This means that to get ΔW=mΔV.
Gravitational potential graphs also exist. On this, the graph shows a bottom right quadrant, as V is -ve. On this graph, the gradient at each point is g, as d(V)/dr=g.
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This would also mean that on a g-r graph, ΔV would be the area under.
Also another equation for g=-ΔV/Δr. In the textbook this is specified to be over small distances, which makes sense as the actual relationship between g and V is through differentiation/ integration, not a common multiple.
Now onto Equipotentials. Picture an object, like the earth, and then you draw circles around it increasing in radius (so these lines are perpendicular to the field lines at each point). Each of these lines are circular, so the radius is the same all around it. This means that along a particular line g and V are the same. This explains the name, as along the line there is a constant gravitational potential, or all points have equal potential (equipotential). This means that if you move an object along an equipotential, so the distance from the other object is constant, no work is done.
So the last major section we look at is the orbits of planets and satellites.
Fun fact: did you know that time period squared is proportional to radius cubed (T² ∝ r³), called Keplar's 3rd law (though we don't care about the name). This is something mentioned on the spec, and talks about objects in orbit around another object, e.g. a satellite out waiting for a planet, so it involves circular motion and gravitational fields. First you set (GMm)/r²=mv²/r, as the gravitational F=centripetal F. Also the m cancel outs, so the circular motion doesn't depend on the orbiting object, only the larger one. You can rearrange to v²=GM/r. You might notice this is similar to the V equation, but please don't notice that, as it is confusing and I don't think we need to worry about that too much. So from circular motion v=2πr/T, and subbing that into the other rearranged equation gives 4π²r²/T²=GM/r, then rearrange to T²=(4π²/GM)r³, which shows (T² ∝ r³) with a constant of proportionality of 4π²/GM. Because of this if a T²-r³ graph was drawn (T² on Y, r³ on X), 4π²/GM would be the gradient. The spec also says you could plot a log graph for this, which I don't know why you would, but it would just be laws of logs.
Synchronous orbit, means that the orbiting object orbits at the same rate that the larger object spins, e.g. a satellite stays at the same position above a place on Earth, e.g. if it was beaming a death laser onto a specific city, it would need to spin around at the same rate the earth spins, or the laser would end up zapping a neighbouring cities as well. For Earth, this means that the time period is 24hours, so you can use the equation above to find out the height the satellite needs to be. The word to describe this would be Geostationary orbit, which describes synchronous orbit specifically for Earth. The r turns out being around 42 Mm (42,000km), which is 36Mm above the surface of Earth (might be a number to remember for a multiple choice Q)
A satellite will have kinetic and potential energy, and the total energy is Eₖ+Eₚ. However, Eₚ is -ve so actually decreases the total energy, ½mv²-GMm/r. From above v²=GM/r (centripetal F=Grav F), so Eₜ=½mGM/r-GMm/r = -½mGM/r = GMm/2r
Last but not least is escape velocity, which is the speed an object needs to be going to escape gravity. This means the kinetic energy needs to overcome potential energy, so for the minimum necessary kinetic energy, Eₖ=Eₚ, which rearranges to v=√2GM/r. (We use GMm/r as Eₚ , so it isn't -ve as the v will be going up, in the opposite direction to Eₚ, so it would be -- so +ve). You can also sub in g to get v=√2gr.
And it looks like we might have made it, yes, it looks like we've made it to the end. This topic wasn't the worst worst, but it is unfortunately going to get a lot worse in fields. Happy gravity!
