1.

Two satellites of earth, S_(1) and S_(2) are moving in the same orbit. The mass of S_(1) is four times the mass of S_(2). Which one of the following stetements is true ?

Answer»

The potential energies fo earth and satellite in the two cases are equal.
`S_(1)` and `S_(2)` are moving with the same speed.
The kinetic energies of the two satellites are equal.
The time period of `S_(1)` is four times that of `S_(2)`.

Solution :The satellite of mass m is moving in a circular orbit of radius r.
`:.` Kinetic ENERGY of a satellite, `K=(GM m)/(2r )""`..(i)
Potential energy of a satellite, `U=(-GM m)/(r )""`.. (ii)
Orbit speed of satellite , `v=sqrt((GM)/(r ))""`...(iii)
Time-period of satellite, `T=[(((4pi^(2)))/(GM))r^(3)]^(1//2)""`...(iv)
GIVEN `m_(S_(1))=4_(m_(S_(2))`
Since M, r is same for both the satellites `S_(1)` and `S_(2)`.
`:.` From equation (ii), we get `U PROP m`
`:. (U_(S_(1)))/(U_(S_(2)))=(m_(S_(1)))/(m_(S_(2)))=4" or ",K_(S_(1))=4K_(S_(2))`
HENCE option (c ) is wrong.
From (iv), since T is independent of the mass of a satellite, time period is same for both the satellites `S_(1)` and `S_(2)`. Hence option (d) is is wrong.


Discussion

No Comment Found

Related InterviewSolutions