Physics, asked by mounibeauty1710, 9 months ago

The magnitude of gravitational field at a distance 3/2a from the centre

Answers

Answered by rohitsharma2k613
0

Answer:

This chapter deals with the calculation of gravitational fields and potentials in the vicinity of various shapes and sizes of massive bodies.  The reader who has studied electrostatics will recognize that this is all just a repeat of what he or she already knows.  After all, the force of repulsion between two electric charges q1 and q2 a distance r apart in vacuo is  20214rqqπε, where ε0 is the permittivity of free space, and the attractive force between two masses M1and M2 a distancer apart is 221rMGM, where G is the gravitational constant, or, phrased another way, the repulsive force is 221rMGM−. Thus all the equations for the fields and potentials in gravitational problems are the same as  the  corresponding  equations  in  electrostatics  problems,  provided  that  the  charges  are replaced by masses and 4πε0 is replaced by −1/G. I  can,  however,  think  of  two  differences.    In  the  electrostatics  case,  we  have  the possibility of both positive and negative charges.  As far as I know, only positive masses exist.    This  means,  among  other  things,  that  we  do not  have  “gravitational  dipoles”  and all the phenomena associated with polarization that we have in electrostatics. The second difference is this.  If a particle of mass m and charge q is placed in an electric field E,  it  will  experience  a  force qE,  and  it  will  accelerate  at  a  rate  and  in  a  direction given by qE/m.   If the same particle is placed in a gravitational field g, it will experience a force mg and an acceleration mg/m  =  g, irrespective of its mass or of its charge.  All masses and all charges in the same gravitational field accelerate at the same rate.  This is not so in the case of an electric field. I have some sympathy  for the idea of introducing a “rationalized” gravitational constant Γ, given by Γ  =  1/(4πG), in which case the gravitational formulas would look even more like  the  SI  (rationalized  MKSA)  electrostatics  formulas,  with  4π  appearing  in  problems with  spherical  symmetry,  2π  in  problems  with  cylindrical  symmetry,  and  no π  in

2problems involving uniform fields.  This is unlikely to happen, so I do not pursue the idea further here

Explanation:

Answered by shanmukhr059
0

Explanation:

you are yh44g give us the same picture as a ni an all day long term for the same thing but I am a lot of the earth will not have to be a body of earth=6400 to get rid my friends in a lot and I will 3rd day 2 year

Similar questions