Saved Bookmarks
| 1. |
Prove that gravitational field intensity at any point in equal to acceleration experienced at that point. |
|
Answer» Solution :The gravitational FIELD intensity `vec(E)_(1)` (here after called as gravitational field) at a point which is at a distance r from `m_(1)` is defined as the gravitational force experienced by unit MASS placed at that point. It is GIVEN by the ratio `(vec(F)_21)/(m_2)` (where `m_(2)` is the mass of the object on which `vec(F)_(21)` acts) `vec(F)_(21)=-(Gm_(1)m_(2))/(r^2)hat(r)""...(1)` Using `vec(E)_(1)=(vec(F)_(21))/(m_2)` (1) we get, `vec(E)_(1)=-(Gm_1)/(r^2)hat(r)""...(2)` `vec(E)_(1)` is a vector quantity that points towards the mass `m_(1)` and is independent of mass `m_(2)`. The value of `m_(2)` is taken to be of unit magnitude. The unit `hat(r)` is along the line between `m_(1)` and the point in question. The field `vec(E)_(1)` is due to the mass `m_(1)`. In general, the gravitational field intensity due to a mass M ATA distance r is given by `vec(E)=-(GM)/(r^2)hat(r)""...(3)` In the region of this gravitational field, a mass 'm' is placed at a point P, when mass 'm' interacts with the field `vec(E)` and experiences an attractive force due to M. The gravitational force experienced by 'm' due to 'M' is given by  `vecF_(m) = m vecE` Now we can equate this with Newton's second law `vec(F)=mvec(a)` `mvec(a)=mvec(E)` `vec(a)=vec(E)` |
|