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The period of revolution of planet A around the sun is 8 times that of B. The distance of A from the sun is how many times greater than that of B from the sun.
$ {T_A \over T_B } = \left( {r_A \over r_B } \right)^{3/2} \Rightarrow 8 = \left( {4_A \over 4_B} \right) ^{3/2} \Rightarrow r_A = 4_B (8)^{3/2} = 4r_B $
The earth revolves round the sun in one year. If distace between then becomes double the new period will be ……………...years.
$ {T_2 \over T_1} = \left( { r_2 \over r_1} \right) ^{3/2} = (2) ^{3/2} = 2 \sqrt 2 \Rightarrow T_2 = 2 \sqrt 2 years $
The period of moon’s rotation around th earth is nearly 29 days. If moon’s mass were 2 fold its present value and all other things remained unchanged the period of moons’s rotation would be nearly days
Time period does not depends upon the mass of satellite
If the velocity of planet is given by $ \upsilon = G^a M^b R^c $ then
$ \upsilon = \sqrt { GM \over R } = G^{1/2 } M^{1/2} R^{-1/2} $
What does not change in the field of central force
For central force toraqe is zero $ \therefore \tau = { dL \over dt } = 0 \Rightarrow L = constant $
Suppose the gravitational force varies inversely as the nth power of distance the time period of planet in circular orbit of radius R around the sun will be proportional to
$ m \omega ^2 R \alpha {1 \over R^n } \Rightarrow m \left( {4 \pi ^2 \over T^2} \right) $ $ R \alpha {1 \over R^n } \Rightarrow T^2 \alpha R^{n+1 } $ $ \therefore T \alpha R^ { n+1 \over 2 } $
If the radius of the earth were to shrink by 1% its mass remaing the same, the accelration due to gravity on the earth’s surface would
$ g = { GM \over R^2 } $ is mass remains constant then $ \rho \alpha {1 \over R^2 } $
A body of mass m is taken from earth surface to the height h equal to radius of earth, the increase in potential energy will be
$ \triangle U = { mgh \over 1 + {h / R} } = {1 \over 2 } mgR $
An artificial satellite moving in a circular orbit around earth has a total (kinetic + potential energy) Eo, its potential energy is
P.E. = 2. total energy = $2E_0$ Because we know $ U = -{GMM \over r } and Eo = - { GMM \over 2r } $
Two bodies of masses m1 and m2 are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual agravitational attraction Their relative velocity of apporach at sepration distance r between them is
let velocities of these masses at r distance from each other be v1 and v2 respectively By conservation of momentum $ m_1 V_1 - m_2 V_2 = 0 \Rightarrow m_1 V_1 = m_2 V_2 $....(i) By conservation of energy change in P.E. = change in K.E. $ {Gm_1 m_2 \over r } = { 1 \over 2 } m_1 V_1^2 + {1 \over 2 } m_2 V_2^2 $ $ \Rightarrow {m_1 V_1^2 \over m_1 } + { m_2 V_2^2 \over m_2 } = {2 Gm_1 m_2 \over r } $...(ii) on solving (i) and (ii) $ \upsilon_1 = \sqrt {2Gm_2^2 \over r (m_1 + m_2 } $ and $ \upsilon_2 = \sqrt {2Gm_1^2 \over r (m_1 + m_2 } $ $ Vapp = |V_1| + |V_2| = \sqrt { 2G (m_1 + m_2) \over r } $
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