A uniform solid sphere of radius r = 3.5 km produces a gravitational acceleration of ag on its surface. at what distance from the sphere's center are there points (a) inside and (b) outside the sphere where the gravitational acceleration is ag/ 5?

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aristocles aristocles · Answer 1 · 2018-08-01T04:33:30+02:00

As we know that acceleration due to gravity is given by

[tex]a_g = \frac{GM}{R^2}[/tex]

Part a)

Inside the sphere

[tex]\frac{a_g}{5} = \frac{Gm}{(r)^2}[/tex]

[tex]\frac{GM}{5R^2}= \frac{G\frac{M*\frac{4}{3}\pi r^3}{\frac{4}{3}\pi R^3}}{(r)^2}[/tex]

[tex]\frac{GM}{5R^2}= G\frac{M*r}{R^3}[/tex]

[tex] 5R^2 = \frac{R^3}{r}[/tex]

[tex] r = R/5[/tex]

[tex] r = 0.7 km[/tex]

Part b)

Now for a point outside the sphere the gravity is [tex]a = /frac{a_g}{5}[/tex]

[tex]\frac{a_g}{5} = \frac{GM}{(r)^2}[/tex]

[tex]\frac{GM}{5R^2}= \frac{GM}{(r)^2}[/tex]

[tex](r) = \sqrt5 R[/tex]

[tex]r = (\sqrt5 )*3.5 km[/tex]

[tex]r = 7.83 km[/tex]