Answer
given,
Mass of the solid sphere = 1800 Kg
radius of the sphere,R = 5 m
mass of the small sphere, m = 2.30 Kg
when the Point is outside the sphere the Force between them is equal to
[tex]F = \dfrac{GMm}{r^2}[/tex] when r>R
When Point is inside the Sphere
[tex]F = \dfrac{GMm}{R^2}\ \dfrac{r}{R}[/tex] when r<R
where r is the distance where the point mass is placed form the center
Now Force calculation
a) r = 5.05 m
[tex]F = \dfrac{GMm}{r^2}[/tex][/tex]
[tex]F = \dfrac{6.67\times 10^{-11}\times 1800\times 2.3}{5.05^2}[/tex]
F = 1.082 x 10⁻⁸ N
b) r = 2.65 m
[tex]F = \dfrac{GMm}{R^2}\ \dfrac{r}{R}[/tex]
[tex]F = \dfrac{6.67\times 10^{-11}\times 1800\times 2.3}{5.05^2}\ \dfrac{2.65}{5.05}[/tex]
[tex]F = 5.68\times 10^{-9}\ N[/tex]
