Gravitational field at points on the axis for x - R is - G -0 R3-n - 1-x - R-22 - 8-x2 - i- The value of n is-

Let us calculate the field which would exist if there were no hole. g _ 0 = GM x ^ 2 = 4G 3 x ^ 2 ρ #160;_ 0 π R ^ 3 Also, due to the mass resting in p ...

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Standard XII

Physics

Properties of Charges

Gravitational...

Question

Gravitational field at points on the axis for x > R is $\frac{π G ρ_{0} R^{3}}{n} [\frac{1}{(x - (R / 2))^{2}} - \frac{8}{x^{2}}] ˆ i$ . The value of $n$ is:

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Solution

Let us calculate the field which would exist if there were no hole.

$g_{0} = \frac{G M}{x^{2}} = \frac{4 G}{3 x^{2}} ρ_{0} π R^{3}$

Also, due to the mass resting in place of hole, field generated is

$g^{'} = \frac{G M^{'}}{{(x - R / 2)}^{2}} = \frac{4 G}{3 {(x - R / 2)}^{2}} ρ_{0} π \frac{R^{3}}{8} = \frac{G}{6 {(x - R / 2)}^{2}} ρ_{0} π R^{3}$

Thus,

$g = g_{0} - g^{'} = \frac{π G ρ_{0} R^{3}}{6} (\frac{1}{{(x - R / 2)}^{2}} - \frac{8}{x^{2}}) i$

Answer is $\frac{π G ρ_{0} R^{3}}{6} (\frac{1}{{(x - R / 2)}^{2}} - \frac{8}{x^{2}}) i$

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Gravitational field at points on the axis for x > R is πGρ0R3n[1(x−(R/2))2−8x2]ˆi. The value of n is:

Gravitational field at points on the axis for x > R is $\frac{π G ρ_{0} R^{3}}{n} [\frac{1}{(x - (R / 2))^{2}} - \frac{8}{x^{2}}] ˆ i$ . The value of $n$ is: