To find the height at which the acceleration due to gravity changes from [tex]\(4.6 \ \text{m/s}^2\)[/tex] at the surface to [tex]\(2.25 \ \text{m/s}^2\)[/tex] at some height above the surface, we use the formula for gravitational acceleration at a height [tex]\(h\)[/tex] above the surface of a heavenly body.
The formula for the gravitational acceleration at a height [tex]\(h\)[/tex] from the surface of a body with radius [tex]\(r\)[/tex] is given by:
[tex]\[ g_h = g_s \left(\frac{r}{r + h}\right)^2 \][/tex]
where:
- [tex]\( g_h \)[/tex] is the gravitational acceleration at height [tex]\(h\)[/tex].
- [tex]\( g_s \)[/tex] is the gravitational acceleration at the surface.
- [tex]\( r \)[/tex] is the radius of the heavenly body.
- [tex]\( h \)[/tex] is the height above the surface.
Given:
- [tex]\( g_s = 4.6 \ \text{m/s}^2 \)[/tex]
- [tex]\( g_h = 2.25 \ \text{m/s}^2 \)[/tex]
We need to solve for [tex]\( h \)[/tex]. First, rearrange the formula to solve for [tex]\( h \)[/tex]:
[tex]\[ \frac{g_h}{g_s} = \left(\frac{r}{r + h}\right)^2 \][/tex]
Taking the square root of both sides:
[tex]\[ \sqrt{\frac{g_h}{g_s}} = \frac{r}{r + h} \][/tex]
Rearrange to isolate [tex]\( r + h \)[/tex]:
[tex]\[ r + h = \frac{r}{\sqrt{\frac{g_h}{g_s}}} \][/tex]
Simplify the expression:
[tex]\[ r + h = r \sqrt{\frac{g_s}{g_h}} \][/tex]
Solve for [tex]\( h \)[/tex]:
[tex]\[ h = r \sqrt{\frac{g_s}{g_h}} - r \][/tex]
For simplicity, let's assume the radius of the heavenly body [tex]\( r = 1 \ \text{unit} \)[/tex]. Substitute the given values into the equation:
[tex]\[ h = 1 \times \sqrt{\frac{4.6}{2.25}} - 1 \][/tex]
Now calculate the ratio inside the square root:
[tex]\[ \frac{4.6}{2.25} \approx 2.044 \][/tex]
Taking the square root of 2.044:
[tex]\[ \sqrt{2.044} \approx 1.430 \][/tex]
Thus, the height:
[tex]\[ h = 1.430 - 1 \][/tex]
[tex]\[ h \approx 0.430 \][/tex]
Therefore, the height from the surface of the heavenly body where the gravitational acceleration is [tex]\(2.25 \ \text{m/s}^2\)[/tex] is approximately [tex]\(0.430 \ \text{units}\)[/tex].
