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  • The gravitational field, due to the 'left over part' of a uniform sphere (from which a part as. shown, has been 'removed out '),at a very far off point ,P, located as shown,would be

The gravitational field, due to the 'left over part' of a uniform sphere (from which a part as. shown, has been 'removed out '),at a very far off point ,P, located as shown,would be (nearly):

Option: 1

\frac{5}{6}\frac{GM}{x^{2}}


Option: 2

\frac{8}{9}\frac{GM}{x^{2}}


Option: 3

\frac{7}{8}\frac{GM}{x^{2}}


Option: 4

\frac{6}{7}\frac{GM}{x^{2}}


Answers (1)

best_answer

Mass of small part = mass density of big sphere × volume of small part

\text { or, } m=\frac{M}{\frac{4}{3} \pi R^{3}} \times \frac{4}{3} \pi\left(\frac{R}{2}\right)^{3}=\frac{M}{8}

Gravitational field due to rest part = Gravitational field due to big sphere − Gravitational of small part
\begin{array}{l} =\frac{G M}{x^{2}}-\frac{G M}{\left(\frac{R}{2}+x\right)^{2}} \\ =\frac{G M}{x^{2}}-\frac{G M}{8\left(\frac{R}{2}+x\right)^{2}} \\ =G M\left[\frac{1}{x^{2}}-\frac{1}{8\left(\frac{R}{2}+x\right)^{2}}\right] \end{array}

\\ For \ very \ far \ point \ X>>R/2 \\ \\ So, (R/2+X)=x
 

Gravitational field due to rest part

=G M\left[\frac{1}{x^{2}}-\frac{1}{8 x^{2}}\right]=\frac{7 G M}{8 x^{2}}

Posted by

Ritika Harsh

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