2.32 It is a well known fact that during a total solar eclipse the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.
2.32 It is a well known fact that during a total solar eclipse the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.
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1 Answer
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2.32
Distance of the moon from Earth = 3.84 * 108 m
Distance of the Sun from Earth = 1.496 * 1011 m
Sun's diameter = 1.39 * 109 m
Sun's angular diameter = 1920° = 1920 * 4.85 * 10-6 rad = 9.3 * 10-3 rad
During total Solar eclipse, the moon completely covers the Sun, then the angular diameter of both Sun and the
Moon will be equal
So the angular diameter of moonΘ = 9.3 * 10-3 rad, distance d of moon from Earth = 3.84 * 108 m and
diameter of the moon = angular diameter * distance
So the approximate diameter of the Moon = 9.3 * 10-3 * 3.84 * 108 m = 35.712 * 105 m
Similar Questions for you
According to question, we can write
Total moles of gas = n = nOxygen + nOxygen =
Volume of gas = 12 × 22.4 litre = 268.8 litre = 2.688 × 105 cm3
Viscosity = Pascal . Second
=
= x = 1, x + 2y = -1, -x + z = 1
y = -1, Z = 0
Viscosity = [P1A-1T0]
1 msD = 1mm
10 vsD = 9msD
1vsD = 0.9 MsD
L.C. = 1MSD – 1VSD = 1 – 0.9 = 0.1 mm
Zero error = 4LC = 0.4 mm
Reading = MSR + VSR + correction
= 4.1 cm + 6 * .01 cm + (-0.04 cm) = (4.1 + 0.06 – 0.04) cm
= 4.12 cm = 412 * 10-2 cm
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