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 (MCQ) for Practice on Gravitation

Equations to be remembered in respect of the section on gravitation were given in the post dated 9^th May 2008. A few multiple choice practice questions were discussed in the post dated 12^th May 2008 followed by a free response practice question on gravitation in the post dated 15^th May 2008. You can access all posts on gravitation on this site by clicking on the label ‘gravitation’ below this post.

Today we will discuss a few more multiple choice practice questions on gravitation. The following questions are meant for AP Physics B as well as C:

(1) The speed of an artificial satellite moving in an orbit of radius r around the earth is increased by 41.4%. The satellite will

(a) continue to move in the same orbit

(b) move in an orbit of radius 1.414 r

(c) move in an orbit of radius 0.414 r

(d) move in an orbit of radius 0.586 r

(e) escape into the outer space

The orbital speed v of any satellite is given by

v =√(gr) where g is the acceleration due to gravity at the orbit of radius r.

[You will get this by equating the centripetal force to the gravitational pull:

mv²/r = mg]

The escape velocity (v_escape) of a body at a point distant r from the centre of the earth where the acceleration due to gravity is equal to g is given by

v_escape =√(2gr)

Thus the escape velocity is √2 times the orbital speed.

The speed v of the satellite becomes 1.414 times the initial value on increasing the speed by 41.4%. Now, 1.414 v = √2 v. Therefore the new speed is the escape speed so that the satellite will escape into the outer space.

(2) A launch vehicle carrying an artificial satellite of mass m is set for launch on the surface of the earth of mass M and radius R. What is the minimum energy to be spent by the launch vehicle on the satellite so that the satellite will move in a circular orbit of radius 7R? (Gravitational constant = G)

(a) GMm /7R

(b) 6GMm /7R

(c) 6GMm /14R

(d) 13GMm /14R

(e) GMm /14R

The satellite of mass m is initially at rest and therefore its initial energy (E_i) is its gravitational potential energy given by

E_i = – GMm/R

When the satellite is in its orbit of radius 7R, its final energy (E_f) is given by

E_f = – GMm/(2×7R) = – GMm/(14R)

[Note that the total energy of a satellite in an orbit of radius r is – GMm/2r]

The minimum energy required for placing the satellite in its orbit is E_f – E_i given by

E_f – E_i = – [GMm/(14R)] – [– GMm/R]

= (GMm/R) – [GMm/(14R)]

= 13GMm /14R

(3) A small body is projected vertically up with a speed equal to half the escape speed from the earth’s surface. If the radius of the earth is R, what is the maximum height (h) reached by the body? (Neglect air resistance).

(a) R/3

(b) R/2

(c) R

(d) 2R

(e) 5R

When the body projected vertically reaches the maximum height, its speed will be zero since its initial kinetic energy us used up in increasing its gravitational potential energy. Therefore we have

½ m (v_e/2)² = [–GMm /(R+h)] – [–GMm /R] where m is the mass of the body, M is the mass of the earth, v_e is the escape speed from the earth’s surface and G is the gravitational constant.

[The first term on the right hand side of the above equation is the gravitational potential energy of the body at height h. The second term is the gravitational potential energy of the body on the surface of the earth].

Since v_e = √(2GM/R), the above equation becomes

GM /4R = (GM /R) –[GM /(R+h)]

Or, GM /4R = GMh /[R(R+h)

Therefore, ¼ = h/(R+h) from which h = R/3

(4) The orbit of a planet moving around the sun is elliptical with the sun S at one focus of the ellipse. If the orbital speeds of the planet while at points A and B (Fig.) are V₁ and V₂ and the distances SA and SB are d₁ and d₂ respectively, V₁/ V₂ is equal to

(a) √(d₁ /d₂)

(b) √(d₂ /d₁)

(c) d₁ /d₂

(d) d₂ /d₁

(e) (d₂ + d₁) /(d₂ – d₁)

To obtain the answer you may apply Kepler’s law of areas (in respect of plnetary motion), which says that the straight line joining the sun to the planet sweeps equal areas in equal intervals of time. The adjoining figure shows path lengths CD and EF traced by the planet in equal intervals (say, ∆t) and hence the areas CADS and EBFS must be equal. If ∆t is sufficiently small, CD and EF will become straight line segments and the areas will become triangular. Equating the areas of these triangles, we have

½ ×CD×AS = ½ ×EF×BS

Or, ½ × V₁ ∆t × d₁ = ½ × V₂ ∆t × d₂

Therefore, V₁/ V₂ = d₂ /d₁

[You can easily get the answer from angular momentum conservation:

m V₁ d₁ = m V₂ d₂

where m is the mass of the planet. This gives V₁/ V₂ = d₂ /d₁]

(5) We know that the gravitational force between two point masses is inversely proportional to the square of the distance between them. If the gravitational force between two point masses were (let us imagine) inversely proportional to the n^th power of the distance between them, what would be the relation between the orbital period (T) of a planet around the sun and the mean distance (r) of the planet from the sun?

(a) T² α r²ⁿ

(b) T² α rⁿ

(c) T² α r²ⁿ⁺¹

(d) T² α r⁽ⁿ⁺¹⁾

(e) T² α r^{(n – 1)}

For simplicity, let us assume that the orbit is circular. (Circular orbit is only a special case of the general elliptical orbit and the relation between T and r will be unchanged). The centripetal force required for the circular motion is supplied by the gravitational pull so that we have

mrω² = GMm/rⁿ where m is the mass of the planet, ω is its orbital angular velocity, M is the mass of the sun and G is the gravitational constant.

The above equation gives ω² = GM/r⁽ⁿ⁺¹⁾

Since ω = T/2π, we have

(T/2π)² = GM/r⁽ⁿ⁺¹⁾

Therefore, T² α r⁽ⁿ⁺¹⁾

You will find some useful multiple choice questions on gravitation (with solution) at physicsplus.