To solve this problem, we need to determine the height from the Earth's surface at which the acceleration due to gravity is 4 m/s².
### Given Data:
- Mass of the Earth, [tex]\( M_{\text{earth}} = 6 \times 10^{24} \)[/tex] kg
- Radius of the Earth, [tex]\( R_{\text{earth}} = 6400 \)[/tex] km
- New gravitational acceleration, [tex]\( g_{\text{new}} = 4 \)[/tex] m/s²
- Gravitational constant, [tex]\( G = 6.67430 \times 10^{-11} \)[/tex] m³ kg⁻¹ s⁻²
### Steps to Solve:
1. Convert the radius of the Earth from kilometers to meters:
[tex]\[
R_{\text{earth}} = 6400 \times 10^3 \text{ meters}
\][/tex]
2. Use the formula for gravitational acceleration at a distance (R + h) from the center of the Earth:
[tex]\[
g_{\text{new}} = \frac{G \cdot M_{\text{earth}}}{(R_{\text{earth}} + h)^2}
\][/tex]
We need to solve for [tex]\( h \)[/tex] (the height above the Earth's surface).
3. Rearrange the formula to find [tex]\( h \)[/tex]:
[tex]\[
(R_{\text{earth}} + h)^2 = \frac{G \cdot M_{\text{earth}}}{g_{\text{new}}}
\][/tex]
[tex]\[
R_{\text{earth}} + h = \sqrt{\frac{G \cdot M_{\text{earth}}}{g_{\text{new}}}}
\][/tex]
[tex]\[
h = \sqrt{\frac{G \cdot M_{\text{earth}}}{g_{\text{new}}}} - R_{\text{earth}}
\][/tex]
4. Calculate the term inside the square root:
[tex]\[
\frac{G \cdot M_{\text{earth}}}{g_{\text{new}}} \approx 1.0011449999999998 \times 10^{14} \text{ m}^2
\][/tex]
5. Take the square root of the term:
[tex]\[
\sqrt{1.0011449999999998 \times 10^{14}} \approx 10005723.362156281 \text{ meters}
\][/tex]
6. Subtract the Earth's radius to find the height [tex]\( h \)[/tex]:
[tex]\[
10005723.362156281 \text{ meters} - 6400 \times 10^3 \text{ meters} \approx 3605723.3621562812 \text{ meters}
\][/tex]
### Conclusion:
The height from the Earth's surface at which the acceleration due to gravity is 4 m/s² is approximately [tex]\( 3.6 \times 10^6 \)[/tex] meters, or 3.6 million meters.
