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A satellite of mass m is orbiting close to the surface of the earth (Radius R = 6400 km) has a K.E. K. The corresponding K.E. of satellite to escape from the earth’s gravitational field is
$ k = { GMm \over 2R }$
A planet moving along an elliptical orbit is closest to the sun at a distance $r_1$ and farthest away at a distance of $r_2$. If $v_1$ and $v_2$ are the liner velocities at these points respectively, then the ratio ${ \upsilon_1 \over \upsilon_2 } $ is
$ v_1 r_1 = v_2 r_2 $ (angular momentum is constant)
A geostationary satellite is orbiting the earth at a height of 5 R above that of surface of the earth. R being the radius of the earth. The time period of another satellite in hours at a height of 2R from the surface of earth is hr
$ { T_1^2 \over T_2^2 } = { R_1^3 \over R_2^3 } = { (6R)^3 \over (3R)^3 } = 8 $ $ \therefore T_2^2 = { 24 \times 24 \over 8 } = 72 $ $ \therefore T_2 = 6 \sqrt 2 $
The period of a satellite in a circular orbit of radius R is T. the period of another satellite in a circular orbit of radius 4R is
$ {T_1 \over T_2 } = \left( {R_1 \over R_2 } \right)^{3/2} = \left( R \over 4R ) \right) ^{3/2} $ $ \therefore T_2 = 8T_1 $
The orbital speed of jupiter is
orbital radius of Jupiter > orbital radius of Earth $ {V_ J \over V_e } = { \gamma _e \over \gamma_j } $ As $ r_j \lt r_e$ therefore $ V_j \lt Ve $
Kepler’s second law regarding constancy of aerial velocity of a planet is consequence of the law of conservation of
$ { dA \over dt} = { L \over 2m } = constant $
The largest and shortest distance of the earth from the sun are $r_1$ and $r_2$
its distance from the sun when it is at the perpendicular to the major axis of
the orbit drawn from the sun
The earth moves around the sun in elliptical path, so by using the properties of ellipse $ r_1 = (1+e ) a \,and \,r^2 = (1-e) r_2 , a = {r_1 + r_2 \over 2 } $ $ \Rightarrow r_1 r_2 = ( 1 -e^2 )a^2 $ where a= semi major axis b= semi minor axis e= eccentircity Now required distance = sem latysrectum = $b^2/9$ $ = a^2 {(1-e^2) \over a } = { r_1 r_2 \over (r_1+r_2) /2 } = { 2r_1r_2 \over r_1 +r_2 } $
A satellite of mass m is circulating around the earth with constant angular velocity. If radius of the orbit is Ro and mass of earth M , the angular momentum about the center of earth is
Angular momentum = $ Mass \times orbital velocity \times Radius$ $ = m \times \sqrt { GM \over R_o } \times R_o$ $ = m \sqrt {GMR_o }$
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 $
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