Question

The gravitational potential is a distance z=0.17 m away from a distribution of mass that has the following equation:

Vsubg= G(M/R^2)(R^2+z^2)^(1/2)
What is the magnitude of the gravitational field (in nN/kg) associated with this distribution of mass at this position? In this equation the variables M and R are constants and equal to 110 kg and 0.55 m, respectively.

Answer 1

gravitational potential is given as

`V = (GM)/(R^2)(R^2 + Z^2)^0.5`

`E = -(dV)/(dz)`

`E = -(GM)/(R^2)(d)/(dz)(R^2+z^2)^0.5`

`E  = -(GM)/(R^2)*0.5(R^2+z^2)^-0.5*2z`

`E = -(GM*z)/(R^2*(R^2 + z^2)^0.5)`

now plug in all values given

M = 110 kg

R = 0.55 m

z = 0.17 m

`E = -(6.67 * 10^(-11)* 110*0.17)/(0.55^2*(0.55^2 + 0.17^2)^0.5)`

`E = 7.16 * 10^(-9) N/kg`

so above is the field intensity