NEET College Predictor
ApplyKnow possible Govt/Private MBBS/BDS Colleges based on your NEET rank
Gravitational field Intensity is considered one the most difficult concept.
Gravitational field due to Point mass is considered one of the most asked concept.
54 Questions around this concept.
A body of mass 60g experiences a gravitational force when placed at a particular point. The magnitude of the gravitational field intensity at that point is:
Two bodies of masses are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is
What is the equation for the gravitational field due to a uniform circular ring of radius 'a' at a point on its axis?
NEET 2024: Cutoff (OBC, SC, ST & General Category)
NEET 2024 Admission Guidance: Personalised | Study Abroad
NEET 2025: Syllabus | Most Scoring concepts | NEET PYQ's (2015-24)
Dependence of intensity of gravitational field (E) of earth with distance (r) from centre of earth is correctly represented by
Starting from the centre of the earth having radius R, the variation of g (acceleration due to gravity) is shown by
The gravitational field intensity due to a uniform solid sphere at a distance r (for outside the solid sphere) from its centre is proportional to:
Gravitational field-
It is space or surrounding in which a material body feels the gravitational force of attraction.
Gravitational field Intensity-
It is the force experienced by a unit mass at a point in the field.
It is denoted by I
If the mass of a body is m then I is given by
It is a vector quantity
If the field is produced by a body M the direction of its Gravitational field Intensity is always towards the center of gravity of M.
Gravitational field due to Point mass
If the point mass M is producing the field and test mass is at distance r as shown in fig
So Force is given as
And the corresponding I is given by
Where
Means As the distance (r) of test mass from point (M) Increases I decreases.
Superposition of Gravitational field
The net Intensity at a given point due to different point masses (M1,M2,M3…) can be calculated by doing the vector sum of their individual intensities
Point of zero intensity-
Let m1 and m2 are separated at a distance d from each other
And P is the point where net Intensity==0
Then P is the point of zero intensity
Let point P is at distance x from m1
Then For point P =0
Then
And
Gravitational field line-
Field line of Isolated mass-
Field lines are radially Inward
As
So we can say that the intensity of the gravitational field at a point P in the field of Isolated mass is equal to the acceleration of test mass placed at that point P.
Properties of Gravitational field line-
The line includes arrows which represent the direction of the gravitational field.
The magnitude of the gravitational field is proportional to the number of field lines crossing a unit area perpendicular to them.
Value of g is larger when lines are close together
While Value of g smaller when they are far apart
The lines never cross
Lines do not form closed loops
For Uniform circular ring
At the center of ring
At a point on its Axis
Where,
The distance of the point P along the Axis of the ring, from its center .
a= radius of the ring
For Uniform disc
Angle with axis
Radius of disc
At the center of the disc
At a point on its axis
For Spherical shell
Radius of shell
Position of Pt.
Mass of spherical shell
Inside the surface
on the surface
Outside the surface
For Uniform solid sphere-
Inside surface
on the surface
Outside surface
"Stay in the loop. Receive exam news, study resources, and expert advice!"