Answer :
Let's solve the problem step-by-step:
1. Understanding the given data:
- The gravitational acceleration at the surface of the Earth ([tex]\( g_{\text{surface}} \)[/tex]) is [tex]\( 9.8 \, \text{m/s}^2 \)[/tex].
- The gravitational acceleration at some height above the Earth's surface ([tex]\( g_{\text{height}} \)[/tex]) is [tex]\( 2.5 \, \text{m/s}^2 \)[/tex].
- The mass of the Earth ([tex]\( M \)[/tex]) is [tex]\( 6 \times 10^{24} \, \text{kg} \)[/tex].
- The radius of the Earth ([tex]\( R_{\text{earth}} \)[/tex]) is [tex]\( 6400 \, \text{km} \)[/tex], which is [tex]\( 6400 \times 10^3 \, \text{m} \)[/tex].
2. Formulating the relationship using the gravitational formula:
The gravitational acceleration at a distance [tex]\( r \)[/tex] from the center of the Earth is given by the formula:
[tex]\[ g = \frac{G \cdot M}{r^2} \][/tex]
where [tex]\( G \)[/tex] is the gravitational constant and [tex]\( M \)[/tex] is the mass of the Earth.
3. Setting up the ratio:
We can use the known gravitational acceleration at the surface and at height:
[tex]\[ g_{\text{surface}} = \frac{G \cdot M}{R_{\text{earth}}^2} \][/tex]
[tex]\[ g_{\text{height}} = \frac{G \cdot M}{r^2} \][/tex]
Taking the ratio of [tex]\( g_{\text{height}} \)[/tex] to [tex]\( g_{\text{surface}} \)[/tex]:
[tex]\[ \frac{g_{\text{height}}}{g_{\text{surface}}} = \frac{\frac{G \cdot M}{r^2}}{\frac{G \cdot M}{R_{\text{earth}}^2}} = \left( \frac{R_{\text{earth}}}{r} \right)^2 \][/tex]
Substitute the given values:
[tex]\[ \frac{2.5}{9.8} = \left( \frac{6400 \times 10^3}{r} \right)^2 \][/tex]
4. Solving for [tex]\( r \)[/tex]:
[tex]\[ \frac{2.5}{9.8} = \left( \frac{6400 \times 10^3}{r} \right)^2 \][/tex]
[tex]\[ \left( \frac{r}{6400 \times 10^3} \right)^2 = \frac{9.8}{2.5} \][/tex]
[tex]\[ r = 6400 \times 10^3 \times \sqrt{\frac{9.8}{2.5}} \][/tex]
From the calculations:
[tex]\[ r \approx 12671353.518862933 \, \text{meters} \][/tex]
5. Calculating the height above the Earth's surface:
The height above the Earth's surface is the total distance from the center of the Earth minus the Earth's radius:
[tex]\[ \text{height}_{\text{above surface}} = r - R_{\text{earth}} \][/tex]
[tex]\[ \text{height}_{\text{above surface}} = 12671353.518862933 \, \text{m} - 6400 \times 10^3 \, \text{m} \][/tex]
[tex]\[ \text{height}_{\text{above surface}} \approx 6271353.518862933 \, \text{meters} \][/tex]
Final Answer:
- The distance from the center of the Earth where the gravitational acceleration is [tex]\( 2.5 \, \text{m/s}^2 \)[/tex] is approximately [tex]\( 12671353.518862933 \, \text{meters} \)[/tex].
- The height above the Earth's surface where the gravitational acceleration becomes [tex]\( 2.5 \, \text{m/s}^2 \)[/tex] is approximately [tex]\( 6271353.518862933 \, \text{meters} \)[/tex] or about [tex]\( 6271.353518862933 \, \text{km} \)[/tex].
1. Understanding the given data:
- The gravitational acceleration at the surface of the Earth ([tex]\( g_{\text{surface}} \)[/tex]) is [tex]\( 9.8 \, \text{m/s}^2 \)[/tex].
- The gravitational acceleration at some height above the Earth's surface ([tex]\( g_{\text{height}} \)[/tex]) is [tex]\( 2.5 \, \text{m/s}^2 \)[/tex].
- The mass of the Earth ([tex]\( M \)[/tex]) is [tex]\( 6 \times 10^{24} \, \text{kg} \)[/tex].
- The radius of the Earth ([tex]\( R_{\text{earth}} \)[/tex]) is [tex]\( 6400 \, \text{km} \)[/tex], which is [tex]\( 6400 \times 10^3 \, \text{m} \)[/tex].
2. Formulating the relationship using the gravitational formula:
The gravitational acceleration at a distance [tex]\( r \)[/tex] from the center of the Earth is given by the formula:
[tex]\[ g = \frac{G \cdot M}{r^2} \][/tex]
where [tex]\( G \)[/tex] is the gravitational constant and [tex]\( M \)[/tex] is the mass of the Earth.
3. Setting up the ratio:
We can use the known gravitational acceleration at the surface and at height:
[tex]\[ g_{\text{surface}} = \frac{G \cdot M}{R_{\text{earth}}^2} \][/tex]
[tex]\[ g_{\text{height}} = \frac{G \cdot M}{r^2} \][/tex]
Taking the ratio of [tex]\( g_{\text{height}} \)[/tex] to [tex]\( g_{\text{surface}} \)[/tex]:
[tex]\[ \frac{g_{\text{height}}}{g_{\text{surface}}} = \frac{\frac{G \cdot M}{r^2}}{\frac{G \cdot M}{R_{\text{earth}}^2}} = \left( \frac{R_{\text{earth}}}{r} \right)^2 \][/tex]
Substitute the given values:
[tex]\[ \frac{2.5}{9.8} = \left( \frac{6400 \times 10^3}{r} \right)^2 \][/tex]
4. Solving for [tex]\( r \)[/tex]:
[tex]\[ \frac{2.5}{9.8} = \left( \frac{6400 \times 10^3}{r} \right)^2 \][/tex]
[tex]\[ \left( \frac{r}{6400 \times 10^3} \right)^2 = \frac{9.8}{2.5} \][/tex]
[tex]\[ r = 6400 \times 10^3 \times \sqrt{\frac{9.8}{2.5}} \][/tex]
From the calculations:
[tex]\[ r \approx 12671353.518862933 \, \text{meters} \][/tex]
5. Calculating the height above the Earth's surface:
The height above the Earth's surface is the total distance from the center of the Earth minus the Earth's radius:
[tex]\[ \text{height}_{\text{above surface}} = r - R_{\text{earth}} \][/tex]
[tex]\[ \text{height}_{\text{above surface}} = 12671353.518862933 \, \text{m} - 6400 \times 10^3 \, \text{m} \][/tex]
[tex]\[ \text{height}_{\text{above surface}} \approx 6271353.518862933 \, \text{meters} \][/tex]
Final Answer:
- The distance from the center of the Earth where the gravitational acceleration is [tex]\( 2.5 \, \text{m/s}^2 \)[/tex] is approximately [tex]\( 12671353.518862933 \, \text{meters} \)[/tex].
- The height above the Earth's surface where the gravitational acceleration becomes [tex]\( 2.5 \, \text{m/s}^2 \)[/tex] is approximately [tex]\( 6271353.518862933 \, \text{meters} \)[/tex] or about [tex]\( 6271.353518862933 \, \text{km} \)[/tex].
