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?

To solve this question, we use the formula for gravitational acceleration \( g \) due to a sphere of mass \( M \) and radius \( R \), which is given by \( g = \frac{G M}{r^2} \), where \( G \) is the gravitational constant and \( r \) is the distance from the sphere's center to the point where the acceleration is being measured.

For a uniform solid sphere, the gravitational acceleration inside the sphere varies linearly with distance from the center because the mass causing the attraction on the inside of the sphere is only the part that is at a lesser radius than the point under consideration. The formula for the gravitational acceleration at a distance \( r \) from the center inside the sphere is \( g_{\text{inside}} = \frac{G M}{R^3} \cdot r \), where \( R \) is the radius of the sphere.

**(a) Points inside the sphere where the gravitational acceleration is \( \frac{a_g}{5} \):**

Given that \( a_g \) is the acceleration at the surface and \( a_{g_{\text{inside}}} \) is \( \frac{a_g}{5} \), we set up the following equation:

\[ \frac{G M}{R^3} \cdot r_{\text{inside}} = \frac{1}{5} \left( \frac{G M}{R^2} \right) \]

Solving for \( r_{\text{inside}} \), we get:

\[ r_{\text{inside}} = \frac{R}{5} \]

Since the sphere's radius \( R \) is given as 3.5 km, the point inside the sphere where the gravitational acceleration is \( \frac{a_g}{5} \) is:

\[ r_{\text{inside}} = \frac{3.5 \text{ km}}{5} = 0.7 \text{ km} \]

**(b) Points outside the sphere where the gravitational acceleration is \( \frac{a_g}{5} \):**

For points outside the sphere, we go back to the formula \( g = \frac{G M}{r^2} \). If the outside point is at a distance \( r_{\text{outside}} \) from the center where acceleration is \( \frac{a_g}{5} \), we have:

\[ \frac{G M}{r_{\text{outside}}^2} = \frac{1}{5} \left( \frac{G M}{R^2} \right) \]

Solving for \( r_{\text{outside}} \), we get:

\[ r_{\text{outside}}^2 = 5 \cdot R^2 \]

\[ r_{\text{outside}} = \sqrt{5} \cdot R \]

\[ r_{\text{outside}} = \sqrt{5} \cdot 3.5 \text{ km} \]

\[ r_{\text{outside}} = \sqrt{5} \cdot 3.5 \text{ km} \approx 7.8 \text{ km} \]

So, the gravitational acceleration will be \( \frac{a_g}{5} \) at a distance of approximately 7.8 km from the center of the sphere.