AP Physics B – Free Response Questions (for Practice) on Thermodynamics

I give below two typical free response practice questions on thermodynamics:

(1) One mole of an ideal mono atomic gas at standard temperature (T₀) and standard pressure (P₀) is contained in a cylinder fitted with a frictionless piston. The gas has volume V₀ under these conditions. The following three operations are performed on the gas:

(i) The gas is heated at constant volume so that its pressure is doubled.

(ii) The gas is then made to undergo isothermal expansion so that the pressure is restored to the initial value.

(iii) The gas then undergoes an isobaric compression so that its volume is restored to the initial value.

Now, answer the following questions:

(a) Draw a P-V diagram to represent the processes numbered (i), (ii) and (iii) above, marking the initial (as well as final) state of the gas by point A, and the intermediate states by points B and C.

(b) What is the final temperature of the gas? Justify your answer.

(c) During the process (i) what is the work done by the gas? Justify your answer.

(d) Calculate the volume (V₂) of the gas at the end of the isothermal expansion mentioned in process (ii).

(e) If the work done by the gas during its isothermal expansion is given to be 1.45P₀V₀, calculate the net work done during the entire cycle of operations involving the processes (i), (ii) and (iii).

(2) Two moles of nitrogen are contained in a vertical conducting cylinder provided with a frictionless piston of negligible mass and area equal to 1.5×10^–2 m² (Fig.). Nitrogen behaves as an ideal gas. It is brought to the following states through a sequence of operations as follows:

(i) Initially the temperature of nitrogen is the same as the laboratory temperature of 27º C and its volume is V₁. This state of the gas is designated as state A

(ii) The piston is now pulled up with a gradually increasing force of final value 60 N, maintaining the temperature at 27º C itself. The volume of the gas now becomes V₂. This state of the gas is designated as state B.

(iii) The temperature of the gas is now increased to 100º C by placing the cylinder in contact with a bath containing boiling water. The volume of the gas now becomes V₃. This state of the gas is designated as state C.

(iv) With the cylinder still in contact with the boiling water bath, the pulling force of 60 N on the piston is now withdrawn and the gas is allowed to get compresed to state D.

(v) The gas in the cylinder is now subjected to a certain process in which the piston moves up through a distance of 0.1 m without any change in the pressure of the gas. The gas attains state E after the expansion. Now answer the following:

(a) Calculate the volume (V₁) of the gas in state A

(b) Indicate by putting a tick (√) mark whether the process from state A to state B is adiabatic, isothermal or isobaric

Adiabatic____ Isothermal_____ Isobaric____

Justify your answer

(c) Calculate the pressure of the gas in state C.

(d) Between which states work is done on the gas? Indicate by tick mark:

A to B____ B to C____ C to D____ D to E____

Justify your answer.

(e) Calculate the work done when the gas changes its state from D to E as mentioned in (v) above.

Try to answer the above questions which carry 10 points each. You can take about 11 minutes for answering each question.

I’ll be back soon with model answers for your benefit.

AP Physics B – Multiple Choice Questions on Thermodynamics

The essential things to be remembered in thermodynamics were given in the post dated 21^st May 2008.

We will now discus some typical multiple choice practice questions related to thermodynamics. Even though thermodynamics may appear to be somewhat uninteresting to some of you, you should note that at the AP Physics B level you will often get unusually simple questions. So, cheer up and go through these questions and after your own trial, see their solution:

(1) In the PV diagram shown, points A and B represent two states of a given mass of an ideal gas. This gas is taken from state A to state B in two different ways:

(i) An isobaric process first and an isochoric (constant volume) process next.

(ii) An isochoric process first and an isobaric process next.

If the work done by the gas in cases (i) and (ii) are respectively W₁ and W₂, pick out the most suitable statement:

(a) W₁> W₂

(b) W₂> W₁

(c) W₁= W₂

(d) W₁= W₂/2

(e) W₁= W₂/3

In both cases the gas expands and therefore work is done by the gas. You should remember that no work is done in the case of an isochoric process. So, no work is done in the processes indicated by PB and AQ (Fig.).

Work is done by the gas during the processes indicated by AP and QB. Since the work done is equal to the area under the PV curve, more work is done during the process AP. Therefore, W₁> W₂.

(2) In the cyclic process on an ideal gas shown in the adjoining PV diagram, what is the net work done on the gas during the cycle?

(a) 2PV

(b) 4PV

(c) – 2PV

(d) – 4PV

(e) zero

This is a simple question of the type often asked. You should note that in a cyclic process the gas is taken through a cycle of operations and brought back to the original state. If the cycle is clockwise as in the present case, work is done by the gas. So the work done on the gas is negative. Further, the area enclosed by the closed curve gives the work done during one cycle.

Therefore, work done on the gas,

W = – area of the rectangle APBQA

= – 2P×2V = – 4PV.

(3) A sample of gas (assumed to be ideal) is adiabatically compressed to have its volume reduced to 20% of its initial volume. If the internal energy of the gas is increased by 200 J, the work done on the gas must be

(a) 40 J

(b) 100 J

(c) 200 J

(d) 400 J

(e) zero

Since the process is adiabatic, no heat energy flows from the surroundings to the gas or from the gas to the surroundings. So, the entire work done in compressing the gas is utilised in increasing the internal energy of the gas. The work done on the gas is therefore equal in value to the increase in internal energy, which is 200 J.

(4) The figure shows the variation of the pressure of a gas with temperature (and not volume) of a given mass of an ideal gas subjected to various processes starting from the state represented by point A. Which graph represents an isochoric process (a process in which volume is constant)?

(a) AB

(b) AC

(c) AD

(d) AE

(e) AF

Some of you may be baffled on seing this simple question since most of you are accustomed to PV diagrams only. The required graph is AB [Option (a)] since the pressure of a given mass of gas is directly proportional to its absolute temperature when its volume is constant (in accordance with Charles law). This type of pressure-volume relationship exists only in the case of the process shown by graph AB.

(5) The volume of a gaseous sample is reduced from V₁ to V₂ in three different ways:

(i) isobaric process (ii) isothermal process and (iii) adiabatic process. Then the work done by the gas is

(a) minimum in the isothermal process

(b) maximum in the isothermal process

(c) minimum in the isobaric process

(d) maximum in the isobaric process

(e) maximum in the adiabatic process

You can easily obtain the answer by referring to the PV diagram showing the three processes (Fig.). Since the area under the PV curve gives the work done, the maximum work is for the isobaric (constant pressure) process indicated by the cuve AB. So the correct option is (d).

The isothermal curve is AC and the adiabatic curve is AD. (Note that the adiabatic curve is steeper than the isothermal curve. The slope of the adiabatic curve is γ times the slope of the isothermal curve where γ is the ratio of specific heats of the gas)

We will discuss more questions in this section in due course

AP Physics B – Thermodynamics –Equations to be Remembered

Essential points to be remembered in respect of kinetic theory of gases were discussed in the post dated 13^th March 2008. It will be useful to have a glance at that post at the moment since you will have to deal with different types of gases when you consider different thermodynamic processes. You can access that post by clicking here

Here are the important points you need to remember in the section ‘thermodynamics’.

(1) Zeroth law of thermodynamics:

If two systems are separately in thermal equilibrium with a third system, then the two systems should be in thermal equilibrium.

(2) First law of thermodynamics:

In the case of any system, the energy (ΔQ) supplied to the system is used partly to increase the internal energy (ΔU) of the system and the rest to do external work (ΔW). Therefore,

ΔQ = ΔU + ΔW

(3) Isothermal process is one in which the temperature of the system is kept constant throughout.

In the case of an isothermal process of an ideal gas, PV = constant. (This is Boyle’s law)

If an ideal gas undergoes an isothermal change (at fixed temperature T) from its initial state (P₁, V₁) to its final state(P₂, V₂), the work done is given by

W = _V1 ∫^V2 PdV = _V1 ∫^V2 (RT/V) dV since PV = RT so that P = RT/V. [We have considered one mole of the gas and hence used the universal gas constant R].

Thus W = RT ln(V₂/V₁)

[Note that work is done only if the volume of the gas changes].

(4) Adiabatic process is one in which no heat flows between the system and the surroundings (ΔQ = 0).

In the case of an adiabatic process of an ideal gas, pressure and volume are related as

PV^γ = constant where γ is the ratio of specific heats of the gas.

Adiabatic expansion of a gas will result in lowering of the temperature of the gas where as adiabatic compression will result in rise in temperature of the gas. The volume and the temperature in an adiabatic change are related as

TV ^(γ–1) = constant

Pressure and temperature in an adiabatic change are related as

P^(1–γ) T ^γ = constant

The work done when one mole of an ideal gas undergoes an adiabatic process is given by

W = [1/(γ–1)][P₁V₁ – P₂V₂]

This can be written as W = [R/(γ–1)][T₁ – T₂] = C_v[T₁ – T₂]

(5) Isobaric process is one in which the pressure is constant. The work done by a gas in an isobaric process is given by

W = P(V₂ – V₁)

(6) Isochoric process is one in which the volume of the gas remains constant. Therefore, no work is done in an isochoric process.

(7) Cyclic process is one in which the system returns to its initial state. The change in internal energy (ΔU) is zero in a cyclic process.

(8) P–V diagram is the graph obtained by plotting the pressure P on the Y-axis and the volume V on the X-axis. Since the work done is ∫PdV, the area under a PV curve gives the work done.

In the case of a cyclic process, the P-V diagram will be a closed curve and the area enclosed by the curve will give the net amount of work done. Note that if the direction of the process as indicated in the P-V diagram of the cyclic process is clockwise, net amount of work is done by the gas. If the direction is anticlockwise, net amount of work is done on the gas.

(9) P–V diagram of a Carnot engine is shown in the adjoining figure.

The curve AB indicates the isothermal expansion of the working substance (ideal gas) at the (higher) source temperature T₁ and the pressure-volume values change from P₁,V₁ to P₂, V₂. The curve BC indicates the adiabatic expansion of the working substance and its state changes from P₂, V₂, T₁ (at B) to P₃, V₃, T₂ (at C).

The curve CD indicates the isothermal compression of the working substance (ideal gas) at the (lower) sink temperature T₂ and the pressure-volume values change from P₃, V₃ to P₄, V₄. The curve DA indicates the adiabatic compression of the working substance and its state changes from P₄, V₄, T₂ (at D) to P₁, V₁, T₁ (at A).

The net amount of work done by the working substance is equal to the area ABCDA

(10) Efficiency (η) of Carnot engine is given by

η = (Q₁ – Q₂)/Q₁ = (T₁ – T₂)/T₁ where Q₁ is the quantity of heat absorbed from the source at the source temperature T₁ and Q₂ is the quantity of heat liberated to the sink at the sink temperature T₂.

The maximum possible efficiency of one (100%) can be obtained only if the temperature of the sink is 0 K (which cannot be attained).

[Note that the temperatures in the above expression should be in degrees Kelvin].

(11) Coefficient of performance (β) of refrigerator is given by

β = Heat removed from the cold reservoir/ Work done by the pump

Note that a refrigerator is a heat engine operated in the reverse manner. The coefficient of performance is therefore given by

β = Q₂/ W = Q₂/ (Q₁ – Q₂) = T₂/(T₁ – T₂) where Q₂ is the heat taken from the cold reservoir, Q₁ is the higher amount of heat rejected to the surroundings by doing work W (by satisfying the equation, Q₁ = Q₂ + W), T₂ is the temperature of the cold reservoir and T₁ is the temperature of the surroundings.

An ideal refrigerator will have coefficient of performance equal to infinity (which is impossible since heat will have to be transferred from cold body to hot surroundings without the help of external energy source, in violation of the second law of thermodynamics).

We will discuss questions in this section in the next post.

Meanwhile, find some useful multiple choice questions at the following locations:
physicsplus: Heat engine and Refrigerator

physicsplus: Questions on Heat & Thermodynamics

AP Physics B & C –Answer to Free-Response Pracice Question on Gravitation

In the post dated 15^th May 2008, the following free-response question for practice was given to you:

Two identical spherical masses M and M initially at rest at very large separation start to move towards each other under their mutual gravitational attraction. Assume that there are no other appreciable fields to affect their motion.

(a) Show that their relative velocity of approach when they are a distance r apart is √(4GM/r) where G is the gravitational constant.

(b) What is the velocity of the centre of mass of this two mass system?

(c) If the two masses are so large that they could eventually form a binary star system, and move along a circle of radius r₀ with the masses at opposite ends of a diameter, calculate the orbital velocity of each star in terms of G, M and r₀

(d) If there are three identical masses A, B and C (instead of two) forming a ternary star system with the masses at the vertices of an equilateral triangle of side L and revolving along a circular orbit circumscribing the triangle as shown in the figure, calculate the orbital velocity of each star. [Many ternary star systems have been observed using modern telescopes; but, three identical stars with equal separation arranged on the same circle is rare].

(e) What is the gravitational field at the centroid of the triangle? Justify your answer.

As promised, here is a model answer for your benefit.

(a) Since the masses are initially at rest and are separated by a very large distance, their kinetic energy as well as potential energy is zero. When they start moving towards each other, their kinetic energy increase at the cost of their potential energy which becomes more and more negative. When the separation is r, the kinetic energy of the system is ½ MV² + ½ MV² = MV² where V is the magnitude of the velocity of each mass. (Their velocities are V and –V since they are moving towards each other).

Since the gravitational potential energy of the system at separation r is – GMM/r, we have

– GMM/r + MV² = 0 [Total energy is always zero since initially it is zero and there are no external forces].

From the above equation, V = √(GM/r).

The relative velocity of approach is V– (–V) = 2V = √(4GM/r).

(b) The centre of mass of a system can be affected by external forces only. The gravitational attraction between the two masses is an internal force and hence the initial velocity of the centre of mass will be retained. Since the two masses are initially at rest, the centre of mass also is initially at rest. Even though the two masses move towards each other because of the gravitational attraction between them, the velocity of the centre of mass will be zero.

(c) The gravitational attractive force between the masses is GM²/(2r₀)² = GM²/4r₀².

This supplies the centripetal force required for the circular motion so that we have

GM²/4r₀² = MV²/r₀.

Therefore, the orbital velocity V = √(GM/4r₀).

(d) Each mass experiences two forces of equal magnitude F = GM²/L² because of the gravitational pull by the other two masses. The angle between these forces is 60º so that the net force is √(F ² + F ² + 2 FF cos 60º) = F ×√3 = (GM²/L²)×√3.

This resultant force on each mass points toward the centre of the circle whose radius (R) is given by

R = (2/3)L sin60º = L/√3 (Note that the circumcentre coincides with the centroid in the case of equilateral triangle).

Since the resultant gravitational attractive force on each mass supplies the centripetal force required for the circular motion, we have

(GM²/L²)×√3 = MV²/R = MV²/( L/√3).

From this V = √(GM/L).

(e) The gravitational field at the centroid of the triangle is the force on unit mass. But the three equal masses A, B and C exert forces of the same magnitude with angle 120º between their lines of action so that the net force on unit mass placed at the centroid is zero. So, the field is zero.

AP Physics B & C –A Free-Response Practice Question on Gravitation

Working out as many questions as possible will boost your confidence in facing the AP Physics Exam. Constant practice has no substitute and you will definitely be rewarded for your effort. So, start preparing for the AP Physics 2009 Examination without any delay.

The following free response practice question involving gravitation will be useful for both AP Physics B and C aspirants:

Two identical spherical masses M and M initially at rest at very large separation start to move towards each other under their mutual gravitational attraction. Assume that there are no other appreciable fields to affect their motion.

(a) Show that their relative velocity of approach when they are a distance r apart is √(4GM/r) where G is the gravitational constant.

(b) What is the velocity of the centre of mass of this two mass system?

(c) If the two masses are so large that they could eventually form a binary star system, and move along a circle of radius r₀ with the masses at opposite ends of a diameter, calculate the orbital velocity of each star in terms of G, M and r₀ .

(d) If there are three identical masses A, B and C (instead of two) forming a ternary star system with the masses at the vertices of an equilateral triangle of side L and revolving along a circular orbit circumscribing the triangle as shown in the figure, calculate the orbital velocity of each star. [Many ternary star systems have been observed using modern telescopes; but, three identical stars with equal separation arranged on the same circle is rare].

(e) What is the gravitational field at the centroid of the triangle? Justify your answer.

The above question carries 15 points (4+1+3+5+2).

Try to answer this question. I’ll be back with a model answer for you shortly.

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.