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Asked by swayamagarwal2114 | 18 Aug, 2022, 09:04: AM
Expert Answer
When the satellite revolving around the planet in an orbit of radius R , then we have
..................................... (1)
LHS of above expression is centrpetal force and RHS of above expression
is gravitational force of attraction.
where m is mass of stellite, M is mass of planet, R is radius of orbit , v is speed of stellite
in the orbot and G is universal gravitational constant .
if T is period of revolution then we have
v = ( 2 π R ) / T .............................................(2)
By substituting v from eqn.(2) , we rewrite eqn.(1) after simplification as
.................................(3)
RHS of above expression is acceleration of satellite towards planet when satellite is orbiting at radius R
when the planet is completely stopped , it does not have speed in tangetial direction .
Hence due to to gravitational force of attraction it falls towards planet.
Since the force of attraction on satellite depends on distance between satellite and planet ,
acceleration of falling satellite depends on distance r between satellite and planet.
When satellite is at distance r from planet , its acceleration is given as
.....................(4)
where v is the velocity ar distance r . Eqn.(4) can be written as
By integrating both sides of above expression , we get
Let us integrate above expression on both sides
where t is time taken by satellite to reach the planet
we get from above integration
..............................(5 )
Using eqn.(3) , above eqn.(5) can be written as
Hence time t is simplified as
Answered by Thiyagarajan K | 18 Aug, 2022, 02:46: PM
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