Problem Based Learning in
Basic Physics - IX
A. K .Mody
C-14 Beverly Hills, Lam Road, Devlali, Nashik 422401, Maharashtra
H. C. Pradhan
HBCSE,
TIFR, V. N. Purav Marg, Mankhurd
Mumbai
– 400 088
[In this article- ninth in the series of articles we
present problems for a problem based learning course based on order of
magnitude, approximations and errors.]
In this article 9th in the PBL series we
consider various examples from different areas of physics where very often we
make approximations. We are trying to discuss meaning of these approximations.
Approximations with Functions:
For all of the above approximations to be
valid, we must have x << 1.
The limit of approximation is defined by the
value of the first term which is ignored w.r.t the actual value. For example in
the term x2/2
if smaller than the accuracy we are looking for then the term and higher order
terms can be ignored.
Consider the following examples from Physics
to understand the importance of the above mentioned approximations and order of
magnitude calculations.
1.
We say sin θ ≈ θ for small angle where θ is in radians.
Discuss up to what value of θ is this approximation valid? Check for
same value of θ, if the approximation is valid for cos θ ≈ 1
– θ2/2! .
Discussion:
Let
us consider θ = 30o = π/6 radian = 0.523598 for which sin θ
= 0.5 . Thus for at 30odifference
between sin θ and θ is
about 4%. Thus depending on the accuracy
we are looking for we should say sin θ ≈ θ within that many percent. In this case 2nd
term in the expansion become θ3/6 = 0.024 which along with
higher order terms can be ignored if 4% difference is acceptable. Otherwise
this term is included and we can check for the next term. For example you can
convince yourself that at 10o, sin θ ≈ θ the accuracy is within 0.5%.
Now at 30o, cos θ = 0.866025
and 1 – θ2/2 = 0.862922 . Here the error would be
about 0.35%. Next term θ4/4 = 0.0031. Check similar difference for 10o.
2. Assuming earth to be a perfect sphere of radius 6400 km and
Mt. Everest to have height of 8 km,
find the greatest distance on earth surface from where Mt. Everest will be
visible.
Discussion:
Consider
the diagram shown in the figure below.
Here
h is the height of the mountain and R the radius of the earth. If
a tangent is drawn from the peak of the mountain to the earth’s surface, then S
represents the maximum distance on the earth’s surface from where mountain
would be visible (assuming earth’s surface to be perfect sphere).
Referring
to the figure :
. Considering h
<< R we can get cos θ ≈ 1 – h/R
Comparing
this with approximation cos θ ≈ 1 – θ2/2, we get θ2
= 2h/R = 16/6400 = 1/400
i.e.
θ = 1/20 radian or 9/π degree. This
gives S = R θ =
6400/20 = 320 km.
Note
that here we used approximation
where x = h/R. Here x2 = (1/800)2
and higher order terms can justifiably neglected.
3.
Calculation of g:
When we calculate g near earth
surface, We take
where M = 6×1024kg and R = 6.4×106 m. Why do we ignore effect of Sun and Moon near earth
surface, inspite of the fact that Sun is massive?
Let us check the contribution of each near
earth’s surface.
For Earth
SI units
For Sun near earth
SI units
For Moon near earth
SI units , using the
data given in example 4 below. Thus, contribution of the Sun is 4 orders and
that of moon is 6 orders smaller than that due to earth. Hence for practical
purposes we ignore the effect due to the sun and that due to the moon. However,
when it comes to force on large masses as earth and its ocean, even tiny effect
can prove to make significant difference and due to size of the earth can cause
tides in the ocean, (See Example 4 below).
4.
What causes tides:
We want to understand tides in the ocean
based on Newton’s law of gravitation. Light take 500 sec to travel from the Sun
to the Earth and diameter of earth is 6400 km. From this information, find the
difference in gravitational pull (acceleration) experienced by a body when it
is on the near side of the Sun to that when it is on far side of the Sun. [Given: G = 6.67×10 – 11 Nm2/kg2 and mass of the Sun
= 2×10 30 kg]
Solution:
and
Which gives Δg = 1.011×10 – 6 cm/s2.
Considering mass of the oceans, which is a
fraction of earth, this generates a huge difference in the force causing tides.
Now from mass of the moon = 7.3483×1022 kg and earth-moon distance = 384400 km calculate
similar differential g, (say Δg) at earth due to moon. Considering relative position of earth Sun
and moon, can you estimate magnitude that causes tidal effect on earth on a new
moon day, full moon day and at 1st quarter or 3rd
quarter? Here assume that sun-earth-moon are along the same line on a new moon
and full moon day (not in the same order as you can understand). What can
happen if they align on the same line?
Take radius of Sun to be 7×108 m and that of a black hole to be
, where c is the speed of light. Take average height
of a person to be 1.5 m. Calculate differential g between head and toe
of an average height person on the surface of the earth, sun and black hole of
one solar mass.
This tidal gravity on earth acts on the
ocean which is a huge mass (that of ocean), say 100th of that of
earth. Can you calculate this tidal (differential) force on this mass on earth?
Estimate this tidal (differential) force
that a normal human (average mass 50 kg) would experience (difference of g
between head and toe) on the surface of (i) Sun (ii) 1 solar mass black hole.
[Use calculator and preserve at least 8 digits after decimal for all your
calculations].
5.
Why water vapour at 100oC
burns the skin whereas at room temperature does not:
Discussion:
Average
kinetic energy of a gas molecule is considered to be of the order of kBT
where kB
= 1.38×10 – 23 J/K is the Boltzmann constant and T is the
absolute temperature. Accordingly, water molecule (of water vapour) at room
temperature (300 K) has kinetic energy kBT ≈ 1/40 eV whereas at
400 K, kBT ≈ 1/30 eV. Can you explain why
at 300 K water vapor is harmless whereas at 400 K it feels very hot and can
burn the skin? [Hint: the distribution of number of molecules ΔN in the energy range between E and E
+ ΔE is given by Maxwell-Boltzmann distribution function
. For simplicity you may assume energy at E ≈ 1 eV (energy just sufficient
to cause burn) in the interval ΔE ≈
kBT.] [You should get about 104 times more
molecules having energy 1 eV at 400 K compared to that at 300 K].
6.
Estimation of size of an
atom:
This
means mass of one aluminum atom is
where NA =
6.02×1023/mole is the Avogadro number. This means, volume occupied
by one atom (assuming cubic arrangement) is
This gives a =
2.554×10 – 10 m = 2.554 Ao. Here a can be treated as diameter of
the atom.
Actual
crystal structure of Aluminum is FCC that gives 4 atoms per cubic cell. Which
modifies the calculation to get a = 4.05 Ao. The separation
of atoms in FCC is
which is diameter of
the atom.
Thus
Aluminum atom radius is estimated to be 1.43 Ao. We can safely consider typical atomic radius
to be of the order of 1 Ao.
7.
What can exist inside the
nucleus?
Consider an Aluminum atom which has atomic
weight A = 27. Radius of a nucleus is given by R = R0A1/3. Where R0 = 1.22 Fermi. [1 Fermi = 10 -15 m]. This gives RAl
≈ 4 fm. Any particle, proton, neutron or electron
if inside the nucleus will have uncertainty in position of the order of 4 fm.
From uncertainty principle, this gives
.
For electron this gives speed greater than
that of light. Which means electron will have energy E = pc = 5×10 – 11 J ≈ 47 MeV , whereas for proton and
neutron
. Electrons observed in β-decay
have much less energy and thus cannot exist inside the nucleus as independent
particle. Proton and neutron on the
other hand can exist inside the nucleus.
8.
Discussion of SHM
In case of simple pendulum, the differential
equation is solved for small amplitude taking approximation sin θ ≈ θ .
Thus, expression for period obtain would be approximate within those
appropriate limit.
In case of spring we estimate
assuming spring to
behave as elastic for the extension and compressions produced. For a perfectly
elastic spring potential energy
, where as in reality spring can neither be compressed by
indefinite amount nor extended. Thus the V must have higher order terms
and should have expression of the type
. Here values of a
and b are very very small compared to k and thus corresponding
terms become important only when x is large. Time period is also
accurate within this approximation.
9.
Infinity in Physics:
Examples from Electrostatics and Magnetostatics:
Discussion:
In electrostatic we deal with infinite line
of charge, infinite charge plane, and in magnetism we deal with infinite
straight wire carrying current. In real life we do not have such infinite
objects. Then why do we derive such expressions?
In each of the above mentioned case, we find
an expression for field at some distance from these infinite objects.
Consider infinite plane. For any point
outside this plane, we can have two elements exactly at the same distance from
the foot of the perpendicular, on the plane contributing to the field. The
field contributions from these two elements have two components each. The
components parallel to the plane cancel each other and the one perpendicular to
the plane add up and survive. Thus field will always be perpendicular to the
plane. What if we have finite size plate? Even in this case, if we are
sufficiently close to the plane surface (distance very small compared to the
dimensions of the plate), the result of infinite plane applies. In case of
parallel plate capacitor, if the separation between plates is too small
compared to the size of the plate, field lines are all perpendicular except
near edges. These effects would be not negligible if plate separation is
comparable to the size of the plates.
We encourage readers to construct similar
line of arguments for linear charge distribution and straight wire carrying
current.
10.
Infinity in Exponential equations:
Now,
consider famous equations like
(i)
from radioactive
decay
(ii)
for charging of a
capacitor in an RC circuit
(iii)
for growth of current
in an LR circuit
Here
have dimensions of
time and are called time constants. In
each case we can write time constant as τ and exponential term can be
represented as e - t/τ.
We say
that as t → ∞, e - t/τ
→ 0. What does infinite time mean here?
Consider values of e - t/τ at t = τ, 2τ, 3τ, 4τ,
5τ…
e – 1 = 0.3679,
e –
2 = 0.1353, e – 3 = 0.0498, e – 4 = 0.0183, e – 5 = 0.0067. Thus t
→
5τ is good enough for infinity
at 0.7% accuracy and system would stabilize for all practical purpose as the
measuring instrument may not be sensitive enough to record 0.7% variation.
Otherwise we need to consider higher values of t.
Note: Even when we say ex
→ 1 as x → 0, (since
), the value of x determines the limit of this
approximation.
References:
1. Physics Textbook for class XI and XII,
NCERT New Delhi (2006)
2. French A. P. , Vibrations and Waves, W.W. Norton & Co Inc, NY (1971)
3. Halliday, Resnik and Krane, Physics, 5th Ed. John Wiley &
Sons (2002)
No comments:
Post a Comment