If a mass m is placed in the vicinity of another mass M, It experiences a gravitational force of attraction. However, if these two masses are at a very large distance, the force of attraction is negligible. The gravitation potential is the amount of work done per unit mass in bringing it slowly from infinity to some finite distance from M. Hence, gravitational potential is a state function rather than a path function.
One application of this potential is in finding the escape velocity of a body from earth. As we know the gravitation potential energy associated with mass m on earth surface is −GMmR. The mechanical energy conservation gives Vescape=√2GMR–––––––––––––––––––– or Vescape=√2gR. An inquisitive mind decides to get the result by kinematics, considering a particle following curvilinear path from surface to infinity, gravity changes with height as g′=g(1+hR)−2 or vdvdh=−g(1+hR)−2. Solution of the equation within limits h=0 to h=∞ gives Vescape=√2gR–––––––––––––––––.
For most of the objects, either point masses or objects of finite dimension, the variation of potential in space exhibits symmetric behaviour. In case of spherical objects of uniform density, the locus of equipotential points is a spherical shell of any given radius. Hence, this potential field is symmetric about all three axes passing through the centre of spherical object. Consider a somewhat complicated objects as shown in the adjacent figure.
The figure shows a solid cube of edge length 10 cm. The origin is the centre of cube as shown. Eight spherical cavities are formed in this cube, each having a radius of 1 cm and centers at (±2 cm,±2 cm,±2 cm). This figure shows wide range of equipotential surfaces / curves.
In which of the following ways, will, potential vary from the centre to surface of any spherical cavity?
A body of mass m starts moving upwards very slowly from Earth surface due to action of forces as shown. The force F varies with height y as F=2mg(1–ay) where a is positive constant. For first half of total ascent,
If a mass m is placed in the vicinity of another mass M, It experiences a gravitational force of attraction. However, if these two masses are at a very large distance, the force of attraction is negligible. The gravitation potential is the amount of work done per unit mass in bringing it slowly from infinity to some finite distance from M. Hence, gravitational potential is a state function rather than a path function.
One application of this potential is in finding the escape velocity of a body from earth. As we know the gravitation potential energy associated with mass m on earth surface is −GMmR. The mechanical energy conservation gives Vescape=√2GMR–––––––––––––––––––– or Vescape=√2gR. An inquisitive mind decides to get the result by kinematics, considering a particle following curvilinear path from surface to infinity, gravity changes with height as g′=g(1+hR)−2 or vdvdh=−g(1+hR)−2. Solution of the equation within limits h=0 to h=∞ gives Vescape=√2gR–––––––––––––––––.
For most of the objects, either point masses or objects of finite dimension, the variation of potential in space exhibits symmetric behaviour. In case of spherical objects of uniform density, the locus of equipotential points is a spherical shell of any given radius. Hence, this potential field is symmetric about all three axes passing through the centre of spherical object. Consider a somewhat complicated objects as shown in the adjacent figure.
The figure shows a solid cube of edge length 10 cm. The origin is the centre of cube as shown. Eight spherical cavities are formed in this cube, each having a radius of 1 cm and centers at (±2 cm,±2 cm,±2 cm). This figure shows wide range of equipotential surfaces / curves.
If a particle is thrown with escape velocity (ve) from surface of earth, which of the following is the correct energy equation?