InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
What is the time period of satellite near the earth's surface? (neglect height of the orbit of satellite from the surface of the earth)? |
|
Answer» Solution :The force on the satellite due to earth is given by `F=(GmM)/(R^(2))` M-Mass of earth `=6xx10^(24) kg` m-mass of satellite, R-radius of earth `=6.4xx10^(6) m` Let v be the speed of the satellite `v=(2pi R)/(T) RARR T=(2 pi R)/(v)` Required CENTRIPETAL force is provided to satellite by the gravitational force hence `F_(C )=(MV^(2))/(R)` But `F_(C)=(GMm)/(R^(2))` ACCORDING to Newton.s law of gravitation. i.e., `(GMm)/(R^(2))=(m(2pi R)^(2))/(T^(2)R)` `rArr T^(2) =(4 pi^(2) R^(3))/(GM)`, as mass of the earth (M) and G are constants the value of T depends only on the radius of the earth. `rArr T^(2) alpha R^(3)` Substituting the values of M, R and G in above equation we get, T = 84.75 minutes. Thus the satellite revolving around the earth in a circular path near to the earth.s surface takes 1Hour and 24.7 minutes approximately to complete one revolution around earth. |
|