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प्रश्न
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 = k/R sqrt(r^3/g)`. where k is a dimensionless constant and g is acceleration due to gravity.
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उत्तर
By Kepler's third law,
`T^2 ∝ r^3` ⇒ `T ∝ r^(3/2)`
We know that T is a function of R and g.
Let `T ∝ r^(3/2) R^a g^b`
⇒ `T = kr^(3/2) R^a g^b` ......(i)
Where k is a dimensionless constant of proportionality.
Substituting the dimensions of each term in equation (i), we get
`[M^0L^0T] = k[L]^(3/2) [L]^a [LT^-2]^b`
= `k[L^(a+ b + 3/2 T^-2b)]`
On comparing the powers of same terms, we get
`a + b + 3/2` = 0 ......(ii)
`- a2b` = 1 ⇒ b = `- 1/2` ......(iii)
From equation (ii), we get
`a - 1/2 + 3/2` = 0 ⇒ a = – 1
Substituting the values of a and b in equation (i), we get
`T = kr^(3/2) R^-1 g^(-1/2)`
⇒ `T = k/R sqrt(r^3/g)`
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