AP Physics C - The Concept of Potential Energy - How it Simplifies Problem Solving

The concept of potential energy such as electrostatic potential energy, gravitational potential energy, elastic potential energy and the like usually makes seemingly difficult problems in physics simple to solve. Let us consider an example in electrostatics:

Three point positive charges Q₁, Q₂ and Q₃ are arranged equidistant R from the origin O as shown in the adjoining figure. Another point positive charge q of mass m, initially at rest, is released from the origin O. Derive an expression for the velocity of the point charge when it is far away from the origin.

This question can be made simpler if there is only a single charge Q₁ instead of three charges Q₁, Q₂ and Q₃. The complexity of the question can be increased further if the three charges Q₁, Q₂ and Q₃ are at unequal distances from the origin.

Well, let us come to the question as it is. Since you are asked to determine the velocity of the charge q, you may be tempted to think of the electric field and the force (in fact, the resultant force) acting on the charge and the acceleration it produces. Equations you have often used in kinematics also may come to your mind. But the force and the accelearation are variable in this case and you realize that the method you plan to use does not seem to be workable. (A bright student will not have confusions of this sort and he will proceed in the right direction).

Perhaps you might have started thinking in terms of the electric potential and the potential energy if you were asked to determine the kinetic energy of the charge q when it is far away from the origin O. Electrostatic potential is a scalar quantity (unlike electric field, which is a vector quantity) and you can easily manipulate it. In the present problem, you can easily find the net electrostatic potential (due to the charges Q₁, Q₂ and Q₃) at the origin and hence the electrostatic potential energy of the charge q. The electrostatic potential energy at infinity (far away from the origin) is zero and so the change in the potential energy of the charge q is equal to its potential energy at the origin O. By equating this to ½ mv², you get the required velocity v of the point charge when it is far away from the origin O.

Here is how you will proceed:

The electric potential V at the origin (due to the charges Q₁, Q₂ and Q₃) is given by

V = (1/4πε₀)(Q₁/R + Q₂/R + Q₃/R)

Or, V = (1/4πε₀)[(Q₁+ Q₂+ Q₃)/R]

[The electrostatic potential at any point is the work done by an external agency to bring a unit positive charge from infinity (infinite distance) to the point. Therefore, the electrostatic potential V at a point distant r from a point positive charge Q is given by

V = – _∞∫^r [(1/4πε₀)(Q/r²)]dr

The quantity inside the square bracket is the electrostatic force acting on the unit positive charge. When the charge Q is positive, the electric field produced by it and the displacement dr of the unit positive charge are opposite in direction and this is why the sign of the expression for V is negative. The above integral gives

V = (1/4πε₀)(Q/r)

This shows that the potential energy is zero when r = ∞].

The electrostatic potential energy of the charge q when it is at the origin is (1/4πε₀)[(Q₁+ Q₂+ Q₃)q/R].

The loss of potential energy when the charge q moves from the origin O to a point far away from the origin is (1/4πε₀)[(Q₁+ Q₂+ Q₃)q/R] since the potential energy at infinity is zero.

The charge q gains an equivalent kinetic energy ½ mv² and hence we have

½ mv² = (1/4πε₀)[(Q₁+ Q₂+ Q₃)q/R]

The velocity v of the point charge when it is far away from the origin O is therefore given by

v = [(1/2πε₀)(Q₁+ Q₂+ Q₃)q/(mR)]^1/2

* * * * * * * * * * * * * * * * * *

Now consider a simple question from gravitation:

An object of mass m is located at a point P very far away from the moon. The gravitational field of the moon is negligible at the point P and the object is initially at rest. The object is given a gentle push and it moves towards the moon. Determine the speed with which the object will strike the moon’s surface. Assume that the mass and radius of the moon are M and R respectively and the moon’s gravity alone influences the motion of the object.

You can apply the concept of gravitational potential energy to obtain the answer easily as given below:

Initial gravitational potential energy U₁ of the object (at infinite distance from the centre of the moon, which we take as the origin) = 0.

[We have U = – GMm/r where G is gravitational constant. With distance r = ∞, U = 0]

Final gravitational potential energy U₂ of the object at the moon’s surface (at distance R) is given by

U₂ = – GMm/R

The loss of gravitational potential energy = U₁ – U₂ = 0 – (– GMm/R) = GMm/R

This must be equal to the gain in kinetic energy ½ mv² where v is the speed with which the object hits the moon’s surface.

Therefore, ½ mv² = GMm/R, from which v = √(2GM/R)

Note that this is the expression for escape velocity from the surface of the moon. An object at infinity, gently pushed from rest must hit the surface (of the moon or any heavenly body) with the surface value of escape velocity since a body projected from the surface with escape velocity will reach infinite distance before coming to rest.

* * * * * * * * * * * * * * * * *

The above question can almost equally well can be worked out beginning with the concept of force as follows

The gravitational force on the object of mass m at distance r = GMm/r²

Work done dW by the gravitational field in moving the object through a small distance dr along the direction of force is given by

dW = (GMm/r²)dr

We should have put the gravitational force as – GMm/R² since it is directed opposite to the direction of increase of r. Our displacement is from infinity to R and hence dr too is negative. The work dW is indeed positive.

The total work W done by the gravitational field is given by

W = _∞∫^R(GMm/r²)dr = GMm/R

This is equal to the gain in kinetic energy ½ mv². Therefore, ½ mv² = GMm/R, from which v = √(2GM/R)

AP Physics B & C – Multiple Choice Questions (for practice) on Gravitation

The essential points you have to remember in respect of gravitation were discussed in the post dated 9^th May 2008. As promised in the post, we will discuss some multiple choice questions involving gravitation.

(1) A simple pendulum has a time period T when on the earth’s surface. The time period of the same pendulum when it is inside an artificial satellite orbiting around the earth at an altitude equal to the radius of the earth will be

(a) T/4

(b) T/2

(c) 2T

(d) 4T

(e) infinite

In an artificial satellite orbiting the earth, bodies will be weightless. Therefore, there is no restoring force on the bob to oscillate the pendulum. Since the pendulum will not oscillate, its period is infinite [Option (e)].

[To put this in a different manner, the effective value of acceleration due to gravity ‘g’ inside the satellite is zero. On substituting this in the expression for the period, T = 2π√( ℓ/g), you obtain it as infinite].

(2) Infinite number of identical spheres of mass 1 kg each are placed along the X-axis with their centres at x = 1 m, 2 m, 4 m, 8 m, 16 m,…… The magnitude of the resultant gravitational field due to these masses at the origin in terms of the gravitational constant G is

(a) 3G/4

(b) 4G/3

(c) G/2

(d) G/4

(e) infinite

The magnitude of the resultant gravitational field at the origin is given by the sumof the forces on unit mass placed at the origin. These forces being in the same direction, the magnitude of the reultant field (F) is given by

F = G×1/1² + G×1/2² + G×1/4² + G×1/8² +……..

Thus F = G(1 + 1/4 + 1/16 + 1/64 +……..) = 4G/3.

(3) The velocity of escape from the earth’s surface is nearly 11.2 kms^–1. If a body is projected at an angle of 60º with the vertical, its velocity of escape will be

(a) 11.2 kms^–1

(b) 11.2×(2/3) kms^–1

(c) 11.2×(√3/2) kms^–1

(d) 11.2×(2/√3) kms^–1

(e) 11.2/2 kms^–1

The escape velocity is the minimum velocity of projection of a body so as to make it escape into outer space. A body in the gravitational field of the earth has negative gravitational potential energy (equal to –GM/r, with usual notations). By supplying kinetic energy (which is always positive), the total energy is to be made equal to zero to make the body free from the gravitational pull and escape into outer space. So, it is the kinetic energy and hence the magnitude of the velocity of projection that matters and not the direction of projection. The answer therefore is 11.2 kms^–1.

(4) If the radius of the earth were to decrease by 0.1 % without any change in its mass, the acceleration due to gravity on the earth’s surface would

(a) increase by 0.1 %

(b) decrease by 0.1 %

(c) increase by 0.2 %

(d) decrease by 0.2 %

(e) increase by 0.05 %

The expression for the surface value of acceleration due to gravity (g) is

g = GM/R² where where G is the gravitational constant, M is the mass of the earth and R is its radius.

The fractional change in g on changing the quantities on the right hand side is given by δg /g = δG /G + δM /M – 2δR /R

Since G and M have constant values, the first two terms on the right hand side are zero so that δg /g = – 2 δR /R

Therefore, percentage change in g = – 2×(percentage change in R).

The percentage change in the radius R is – 0.1 %. [Negative sign since the change is decrement]

Therefore, percentage change in g = – 2×(– 0.1) = 0.2 %. Since the sign is positive, this is an increment and the correct option is (c).

(5) Imagine a body at rest at height R from the earth’s surface, where R is the radius of the earth. If it falls freely under the gravitational pull of the earth, what will be its velocity just before it hits the earth’s surface where the acceleration due to gravity is g? Neglect the air resistance for the sake of simplicity of the problem.[In a real situation you cannot neglect the air resistance which may even burn the entire body before it reaches the ground!]

(a) √(gR)

(b) √(gR)

(c) √(gR/2)

(d) √(gR/3)

(e) √(2gR/3)

If the height ‘h’ is negligible compared to the radius of the earth, the velocity on hitting the ground would be √(2gh) which you obtain by writing ½ mv² = mgh . But you cannot replace h with R since the value of acceleration due to gravity changes appreciably over the path of the body.

When the body is at altitude R, its distance from the centre of the earth is 2R and its gravitational potential energy is – GMm/2R where m is the mass of the body. On falling down, its kinetic energy increases and its gravitational potential energy decreases (larger negative value).

At the surface of the earth, its gravitational potential energy is – GMm/R.

Therefore, change in gravitational potential energy = –GMm/R – (–GMm/2R) = – GMm/2R. The negative sign shows that the change is a decrement.

The corresponding increment in the kinetic energy is ½ mv² where ‘v’ is the velocity on hitting the ground.

Therefore we have GMm/2R = ½ mv² from which v = √(GM/R) = √(gR) since g = GM/R²

(6) Two planets have radii R₁ and R₂ and mean densities d₁ and d₂ respectively. The ratio of the accelerations due to gravity on their surfaces is

(a) R₁²d₁ : R₂²d₂

(b) R₁d₁ : R₂d₂

(c) R₁d₂ : R₂d₁

(d) R₁d₂² : R₂d₁²

(e) R₁³d₁ : R₂³d₂

The surface value of acceleration due to gravity (g) is given by

g = GM /R²

Since M = (4/3) πR³d where d is the mean density of the planet, we have

g = G×(4/3) πR³d /R² = G×(4/3) πRd

Therefore, g is directly proportional to the product Rd so that the required ratio is R₁d₁ : R₂d₂ [Option (b)].

We will discuss more questions on gravitation in due course.

We have to do the best we can. This is our sacred
human responsibility. 

– Albert Einstein 

AP Physics B & C –Gravitation –Equations to be Remembered

For obtaining good score in gravitation you should remember the following:

(1) The gravitational attractive force F between two point masses m₁ and m₂ separated by a distance r is given by

F = G m₁m₂/r² where G is the gravitational constant (or, constant of gravitation).

If the masses are homogeneous spheres the above equation still holds; but the distance between the masses is to be taken as the distance between the centres of the spheres.

[The gravitational force on a point mass m₂ due to another point mass m₁ should be strictly written as F = –G m₁m₂/r² since the force is attractive and hence is directed opposite to the vector distance r from m₁ to m₂].

The gravitational field produced by a point mass m at a point P distant r from it is the gravitational force acting on unit mass placed at the point P and is equal to Gm/r².

(2) Since the weight of a body of mass m is the gravitational force with which the earth of mass M pulls the body, we have

mg’ = GMm /r² where r is the distance of the body from the centre of the earth and g’ is the acceleration due to gravity at the distance r.

Therefore, the acceleration due to gravity (g’) at a distance r from the centre of the earth is given by

g’ = GM /r²

If the body is on the surface of the earth of radius R, we have the surface value of acceleration due to gravity (g) given by

g = GM /R²

Note that the surface value of acceleration due to gravity depends to a small extent on the latitude λ because of the spin motion of the earth and is given by

g_λ = g – ω²R cos²λ

[g_λ is the surface value of acceleration due to gravity at latitude λ, g is the gravitational acceleration at the poles where the effect of the spin of the earth is absent, ω is the spin angular velocity of the earth and R is the radius of the earth].

Since the value of λ is zero at the equator and 90º at the poles, the surface value of acceleration due to gravity is minimum at the equator and maximum at the poles.

(3) Acceleration due to gravity (g’) at a height ‘h’ is given by g’ = GM/(R+h)² since r = R+h.

If ‘h’ is small compared to the radius ‘R’ of the earth, g’ = g(1–2h/R)

(4) Acceleration due to gravity (g’’) at a depth ‘d’ is given by g’’ = g (1–d/R)

Note that this is true for all values of ‘d’.

(5) Gravitational potential energy (U) of a mass ‘m’ at a height ‘h’ is given by

U= –GMm/(R+h)

This can be written as U = –GMm/r where ‘r’ is the distance from the centre of the earth.

The gravitational potential energy of a mass m₂ in the gravitational field produced by a mass m₁ is given by

U = –Gm₁m₂/r where r is the distance between centres of the masses.

You can as well say that the above expression is the gravitational potential energy of a mass m₁ in the gravitational field produced by a mass m₂ or it is the gravitational potential energy of the pair consisting of the masses m₁ and m₂.

Three masses can make three independent pairs and there will be three terms in the expression for potential energy. Four masses will produce six independent pairs and hence there will be six terms in the expression for potential energy and so on.

(5) Escape velocity (ve) from the surface of earth (or any planet or star) is given by

ve = √(2GM/R) = √(2gR)

Escape velocity from a height ‘h’ = √[2GM/(R+h)] = √[2g’(R+h] = √(2g’r)

Here g’ is the acceleration due to gravity at distance r (= R+h) from the centre of the earth.

(6) Kinetic energy (K.E.) and total energy of a satellite are equal in magnitude. But K.E. is positive where as total energy is negative. The potential energy of a satellite is negative and is equal to twice the total energy. (Note that this is true in all cases of central field motion under inverse square law force, as for example, in the case of the electron in the hydrogen atom).

In the case of a satellite of mass ‘m’ in an orbit of radius ‘r’:

Potential energy = –GMm/r

Kinetic energy = +GMm/2r

Total energy = –GMm/2r

(7) The square of the orbital period of a satellite is directly proportional to the cube of the mean distance from the centre of the earth. (This is in accordance with Kepler’s 3^rd law which states that the square of the orbital period of any planet about the sun is directly proportional to the cube of the mean distance from the centre of the sun).

Therefore, T² α r³

In the case of circular orbits r is the radius of the circle.

In the case of elliptical orbits r is the semi-major axis of the ellipse.

(8) Orbital speed ‘v’ of a satellite in an orbit of radius ‘r’ is obtained by equating the centripetal force required for the motion to the gravitational pull. Thus mv²/r = GMm /r² so that

v = √(GM/r) = √(g’r) where g’ is the acceleration due to gravity at the orbit and M is the mass of the earth (or planet).

(9) The expression for the orbital period of a satellite is

T = 2π√(r³/GM)

[The above expression is obtained from the equation, T =2πr/v].

(10) Since the escape velocity (v_e) and the orbital velocity (v) are given respectively by √(2g’r) and √(g’r), we have v_e = (√2) v

Therefore, if the speed of a satellite orbiting the earth is increased to √2 times its normal orbital speed, it will escape into outer space. In other words, if the speed of a satellite is increased by 41.4%, it will escape into outer space.

In the next post we will discuss questions on gravitation. Meanwhile, find some useful multiple choice questions at physicsplus.