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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.

Respuesta :

gravitational potential is given as

[tex]V = \frac{GM}{R^2}(R^2 + Z^2)^0.5[/tex]

[tex]E = -\frac{dV}{dz}[/tex]

[tex]E = -\frac{GM}{R^2}\frac{d}{dz}(R^2+z^2)^0.5[/tex]

[tex]E  = -\frac{GM}{R^2}*0.5(R^2+z^2)^-0.5*2z[/tex]

[tex]E = -\frac{GM*z}{R^2*(R^2 + z^2)^0.5}[/tex]

now plug in all values given

M = 110 kg

R = 0.55 m

z = 0.17 m

[tex]E = -\frac{6.67 * 10^{-11}* 110*0.17}{0.55^2*(0.55^2 + 0.17^2)^0.5}[/tex]

[tex]E = 7.16 * 10^{-9} N/kg[/tex]

so above is the field intensity

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