Gravitational Potential Energy
What is gravitational potential energy ?
Concept of gravitational potential energy
|Gravitational potential energy|
See from the picture initially both masses are placed at infinity, where gravitational field, gravitational potential as well as gravitational potential energy is assumed zero.
Now think for first mass m1 to bring from infinity to location as shown. What do you think does any work will required to bring mass m1 ? Well no any work required because there is no any mass, which will attract mass m1, So when you will just touch the mass m1 at infinity it will start moving with constant velocity obeying Newton’s first law every body in rest and motion continue in rest and motion unless and until an external force is applied.
When m1 reaches at location shown in figure then just touch to stop it will stop, Hence in this manner no any work done by external agent or gravitational force, So bringing m1 from infinity to m1 location does not required any work.
Now consider for second mass m2, for this case see m1 is already at location shown, which is created its gravitational field around it, So m1 will attract m2 with gravitational force, let at any intermediate position x between infinity to m1.
Hence gravitational force F = Gm1m2/x² this force will increase the velocity of m2 so to avoid increase in velocity you need to apply less little force opposite to this force to move m2 very very slowly, So you have to work in this case.
Now see this force F = Gm1m2/x² , which is variable for different value of x force will be different, so you can not use direct formula of work
W = force.displacement because here force is not constant, So for this you need to take a very small point dx where force will be constant hence you can write.
∫dw = ∫F.dxcos0⁰ why angle cos0⁰ because see the direction of gravitational force and ∞
r r r
w = ∫ Gm1m2/x² dx = Gm1m2∫x⁻²dx = Gm1m2⎣ -1/x ⎤ = -Gm1m2(1/r-1/∞) = -Gm1m2/r
∞ ∞ ∞
Hence work done Wg = -Gm1m2/r , who has done this work this work is done by gravitational force, This work done is stored in the system mass m1 and m2 in the form of gravitational potential energy.
Here negative sign indicate both the masses are now bound with each other with potential energy, If you want to separate them ,then you have to do work.First you need to supply energy Gm1m2/r positive then any one mass will separate from bound system.
So gravitational potential energy is always negative for any masses system or configuration which indicate the system is bounded with potential energy.
You can find gravitational potential energy without integration
Hence Wext = m2(Vr – V∞) = m2(-Gm1/r -0) = -Gm1m2/r , So Wext = -Gm1m2/r this work done is stored in the system in the form of potential energy, This is very powerful formula for work done by external agent calculation.
Same formula is used in electrostatic for work done calculation but basic quantity is charge there So Wext = q∆V.
Important point does this energy belongs to a single mass m1 or m2 answer is no it does not belongs to single mass, This potential energy belongs to system of masses m1 and m2.
What is potential energy mgh ?
Difference of potential energy between two points which is mgh potential energy, you will see here how mgh is calculated ? see the below picture.
|Potential energy mgh|
potential energy at two level system at surface system between Earth and mass m potential energy is -GMem/Re and at height h between Earth and mass m system potential energy -GMem/(Re+h) , can you tell me, which energy value is higher at surface or at height h.
Well at height h level potential energy is high because negative sign associated with energy, So when you go up from the Earth surface there is gain in the potential energy.
Recall i have mentioned above first paragraph potential energy is define as potential energy difference, So the difference in potential energy or gain is given as.
Uf – Ui = ∆U = (-GMem/(Re+h) – (-GMem/Re) )
∆U = Potential energy difference = GMem( 1/Re – 1/(Re+h) ) = GMemh/(Re+h)Re
you recall what is formula (GMe/Re²) this is small g so (GMe/Re²) = g
Hence ∆U = Potential energy difference = mgh
Hence this mgh formula is basically derived from basic potential energy of the system formula (-Gm1m2/r), This negative sign indicate the force is attractive .So the work done is negative.
See below numerical question for potential energy of the system.
In first question you have to find potential energy of equilateral triangle mass shown at vertices 2kg, 4kg and 6kg.
You know potential energy of the system U = -Gm1m2/r in this case three combination is possible (2kg,4kg), (4kg,6kg), (2kg,6kg) So apply the formula.
U = -G2X4/1 + (-G4x6)/1 + (-G2x6)/1 = -8G -24G – 12G = -44G = -44×6.67×10⁻¹¹
U = 293.48x10⁻¹¹ J so this is the potential energy of the triangle system.
Now for second question square find the potential energy of the system.
Can you tell me how many combination will be in square shown, well there will be six combination (1kg,2kg), (2kg,3kg), (3kg,4kg), (1kg,4kg) and two diagonal (1kg,3kg), (2kg,4kg)
Hence total energy of the system will be.
U = -G1x2/1 + (-G2x3)/1 + (-G3x4)/1 + (-G1x4)/1 + (-G1x3)/√2 + (-G2x4)/√2
U = -2G – 6G – 12G -4G – 3G/√2 – 8G/√2 = -24G – 11G/√2 = -24G – 7.8G = -31.8G
U = -31.8×6.67x10⁻¹¹ = -212.1x10⁻¹¹ J is the potential energy of the system.
Remember one formula change in potential energy (∆U) = (-work done by gravitational force) = Work done by external agent .
So ∆U = -Wg = +Wext.
See the above picture and find the external work done.
use the above formula change in potential energy of the system is equal to work done by external agent.
∆U = Wext so Wext = ∆U = Uf – Ui now find final and initial potential energy of the system Uf = -G1x1/2 + (-G1x1)/2 + (-G1x1)/2 = -3G/2
Ui = -G1x1/1 + (-G1x1)/1 + (-G1x1)/1 = -3G
Wext = Uf – Ui = -3G/2 – (-3G) = 3G – 3G/2 = 3G/2 = 1.5×6.67×10⁻¹¹ = 10⁻¹⁰ J
Wext = 10⁻¹⁰ J this work will be done.
Q Find how much work done is required to free a mass m from Earth field from the surface of the Earth. This question is for you.
Dated 13th Jan 2019
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