A spherical cavity made in a solid sphere of radius r. The mass of sphere before the cavity was eked out was m. The gravitational field intensity at the centre of the sphere due to remaining mass is
Answers
Answer:
Explanation:
\displaystyle \bar{E}_1=\dfrac {p\bar{r}}{3\varepsilon_0}=\dfrac {pR}{6\varepsilon _0}.......\left [ \varepsilon _0=\dfrac {1}{4\pi G} \right ]
E
ˉ
1
=
3ε
0
p
r
ˉ
=
6ε
0
pR
.......[ε
0
=
4πG
1
]
Gravitational field at mass mm due to cavity (-p)(−p)
\displaystyle \bar{E}_2=\dfrac {(-p)(R/2)^3 \bar{r}}{3\varepsilon _0R^2}.......\left [ using E=\dfrac {pa^3}{3\varepsilon _0r^2} \right]
E
ˉ
2
=
3ε
0
R
2
(−p)(R/2)
3
r
ˉ
.......[usingE=
3ε
0
r
2
pa
3
]
\displaystyle -\dfrac {(-p)R^3}{24\varepsilon _0R^2}=-\dfrac {-pR}{24\varepsilon _0}−
24ε
0
R
2
(−p)R
3
=−
24ε
0
−pR
Net gravitational field
\displaystyle \bar{E}=\bar{E}_1+\bar{E}_2=\dfrac {pR}{6\varepsilon _0}-\dfrac{pR}{24\varepsilon _0}=\dfrac {pR}{8\varepsilon _0}
E
ˉ
=
E
ˉ
1
+
E
ˉ
2
=
6ε
0
pR
−
24ε
0
pR
=
8ε
0
pR
Net force on m \rightarrow F=m\bar{E}=\dfrac {mpR}{8\varepsilon _0}m→F=m
E
ˉ
=
8ε
0
mpR
Here, \displaystyle p=\dfrac {M}{(4/3)\pi R^3}p=
(4/3)πR
3
M
& \displaystyle \varepsilon_0=\dfrac {1}{4\pi G}ε
0
=
4πG
1
Then, F=\dfrac {3mg}{8}F=
8
3mg