19. The principle of ‘parallax’ in section 2.3.1 is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit ≈ 3 × 1011m. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1” (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1” (second) of arc from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of meters?
Solution:
Diameter of Earth’s orbit = 3 × 1011 m
Radius of Earth’s orbit, r = 1.5 × 1011 m
Let the distance parallax angle be 1'' = 4.847 × 10–6 rad.
Let the distance of the star be D.
Parsec is defined as the distance at which the average radius of the Earth’s orbit subtends an angle of 1''.
∴ We have \(\theta = \frac rD\)
D = \(\frac rD\) = \(\frac {1.5 x 10^{11}}{4.847\times10^{-6}}\) = 0.309 x 10-6 ≈ 3.09 x 1016 m
Hence, 1 parsec ≈ 3.09 x 1016 m
20. The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun?
Solution:
Distance of the star from the solar system = 4.29 ly
1 light year is the distance travelled by light in one year.
1 light year = Speed of light × 1 year
= 3 × 108 × 365 × 24 × 60 × 60 = 94608 × 1011 m
∴ 4.29 ly = 405868.32 × 1011 m
∵ 1 parsec = 3.08 × 1016 m
∴ 4.29 ly = \(\frac {405868.32\times 10^{11}}{3.08\times 10^{16}}\) = 1.32 parsec
Using the relation, \(\theta = \frac dD\)
where.
Diameter of Earth's orbit, d = 3 x 1011m
Distance of the star from the Earth, D = 405868.32 × 1011 m
∴ \(\theta = \frac {3\times 10^{11}}{405868.32 \times 10^{11}}\) = 7.39 x 10-6 rad
But, 1 sec = 4.85 x 10-6 rad
7.39 x 10-6 rad = \(\frac {7.39\times 10^{-4}}{4.85\times10^{-6}}\) = 1.52''
21. Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.
Solution:
It is indeed very true that precise measurements of physical quantities are essential for the development of science. For example, ultra-shot laser pulses (time interval ∼ 10–15 s) are used to measure time intervals in several physical and chemical processes.
X-ray spectroscopy is used to determine the inter-atomic separation or inter-planer spacing.
The development of mass spectrometer makes it possible to measure the mass of atoms precisely.
22. Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity): the total mass of rain-bearing clouds over India during the Monsoon the mass of an elephant the wind speed during a storm the number of strands of hair on your head the number of air molecules in your classroom.
Solution:
During monsoons, a metrologist records about 215 cm of rainfall in India i.e., the height of water column, h = 215 cm = 2.15 m
Area of country, A = 3.3 × 1012 m2
Hence, volume of rain water, V = A × h = 7.09 × 1012 m3
Density of water, ρ = 1 × 103 kg m–3
Hence, mass of rain water = ρ × V = 7.09 × 1015 kg
Hence, the total mass of rain-bearing clouds over India is approximately 7.09 × 1015 kg.
Consider a ship of known base area floating in the sea. Measure its depth in sea (say d1).
Volume of water displaced by the ship, Vb = A d1
Now, move an elephant on the ship and measure the depth of the ship (d2) in this case.
Volume of water displaced by the ship with the elephant on board, Vbe= Ad2
Volume of water displaced by the elephant = Ad2 – Ad1
Density of water = D
Mass of elephant = AD (d2 – d1)
Wind speed during a storm can be measured by an anemometer. As wind blows, it rotates. The rotation made by the anemometer in one second gives the value of wind speed.
Area of the head surface carrying hair = A
With the help of a screw gauge, the diameter and hence, the radius of a hair can be determined. Let it be r.
∴ Area of one hair = πr2
Number of strands of hair = \(\frac {Total\,surface\, area}{Area\, of\, one\, hair\, \pi} = \frac {A}{r^2}\)
Let the volume of the room be V.
One mole of air at NTP occupies 22.4 l i.e., 22.4 × 10–3 m3 volume.
Number of molecules in one mole = 6.023 × 1023
∴ Number of molecules in room of volume V
= \(\frac {6.023 \times 10^{23}}{22.4\times10^{-3}}\times V\) = 134.915 x 1026 V = 1.35 x 1028 V
23. The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107 K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess is correct from the following data: mass of the Sun = 2.0 × 1030 kg, radius of the Sun = 7.0 × 108 m.
Solution:
Mass of the Sun, M = 2.0 × 1030 kg
Radius of the Sun, R = 7.0 × 108 m
Volume of the Sun, V = \(\frac 43\pi r^3\)
= \(\frac 43 \times \frac {22}{7} \times (7.0 \times 10^8)^3\)
= \(\frac {88}{21} \) x 343 x 1024 = 1437.3 x 1024 m3
Density of the Sun = \(\frac {Mass}{Volume}=\frac {2.0 \times 10^{30}}{1437.3\times 10^{24}}\) = 1.4 x 103 kg/m5
The density of the Sun is in the density range of solids and liquids. This high density is attributed to the intense gravitational attraction of the inner layers on the outer layer of the Sun.
24. When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be of arc. Calculate the diameter of Jupiter.
Solution:
Distance of Jupiter from the Earth, D = 824.7 × 106 km = 824.7 × 109 m
Angular diameter = 35.72'' = 35.72 x 4.874 x 10-6 rad
Diameter of Jupiter = d
Using the relation,
\(\theta = \frac dD\)
d = θD = 824.7 x 109 x 35.27 x 4.872 x 10-6
= 143520.76 x 103 = 1.435 x 105 km