An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit of radius r. From Kepler’s Third law about the period of a satellite around a common central body, square of the period of revolution T is proportional to the cube of the radius of the orbit r. Show using dimensional analysis, that
T=
,
where k is a dimensionless constant and g is acceleration due to gravity.
An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit of radius r. From Kepler’s Third law about the period of a satellite around a common central body, square of the period of revolution T is proportional to the cube of the radius of the orbit r. Show using dimensional analysis, that
T= ,
where k is a dimensionless constant and g is acceleration due to gravity.
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1 Answer
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This is a long answer type question as classified in NCERT Exemplar
According to kepler’s law T2 r3
T
T
T
[M0L0T]=K [L]3/2 [L]a [LT-2]b
= k [La+b+3/2T-2b]
On comparing the powers of same terms we get
So a+b+3/2=0
So b=-1/2 and a=-1
T=kr3/2R-1g-1/2
T=
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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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