📖 Samacheer Kalvi · 11th TN - English Medium · Physics Volume 2 · Page 9question

GRAVITATION · Part 11

Chapter 12: Front Matter · Physics Volume 2

from m is defined as the gravitational force experienced by unit mass placed at that point. It is given by the ratio  F (where m is the mass of the object on which  F acts) Using   F in equation ( . ) we get,  Gm =− ( . ) GRAVITATIONAL FIELD AND GRAVITATIONAL POTENTIAL .

. Gravitational field Force is basically due to the interaction between two particles. Depending upon the type of interaction we can have two kinds of forces: Contact forces and Non-contact forces (Figure . ).

Planet Sun Figure . Depiction of contact and non-contact forces Contact forces are the forces applied where one object is in physical contact with the other. The movement of the object is caused by the physical force exerted through the contact between the object and the agent which exerts force. Consider the case of Earth orbiting around the Sun.

Though the Sun and the - - - - Unit Gravitation  E is a vector quantity that points towards the mass m and is independent of mass m , Here m is taken to be of unit magnitude . The unit is along the line between m and the point in question. The field  E is due to the mass m . In general, the gravitational field intensity due to a mass M at a distance r is given by  GM =− ( .

) Now in the region of this gravitational field, a mass ‘m’ is placed at a point P (Figure . ). Mass ‘m’ interacts with the field  E and experiences an attractive force due to M as shown in Figure . .

The gravitational force experienced by ‘m’ due to ‘M’ is given by M F = m E Figure . Gravitational Field intensity measured with an object of unit mass   F mE m = ( . ) Now we can equate this with Newton’s second law   F ma ma mE   ( . )   a ( .

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