Five point masses m each are kept at five vertices of a regular pentagon. Distance of centre of pentagon from any one of the verticle is 'a'. Find gravitational potential and filed strength at centre.
Answers
(1) Gravitational Potential = -5 Gm/a
(2) Field strength at center = 0
Five point masses m each are kept at five vertices of a regular pentagon. Distance of center of pentagon from any one of the vertical is 'a'.
(1) Gravitational Potential due to mass m at distance a is given by ,
- V = - gm/a
- So potential due to all 5 masses will be
- V = -5 Gm/a
(2) Field strength at center :
When the field vectors are added according to the polygon law of vector addition, the sum of the first four vectors will be equal in magnitude and opposite in direction to the fifth vector. Hence they will cancel out.
Hence , Net field will be zero.
( You may refer to the photo attached)
Hence the value of net field strength at the centre is zero.
Explanation:
We are given that:
- Number of point masses = 5
- Distance from pentagon = a
V=5(−Gma)=−5Gma
For E Five vectors of equal magnitudes, when added as per polygon law of vector. Hence, net E is zero .
Hence the value of net field strength at the centre is zero.