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Sagot :
Certainly! Let's work through this problem step-by-step.
We are given several key pieces of information:
1. The gravitational acceleration on the surface of a planet, [tex]\( g_{\text{surface}} \)[/tex], is [tex]\( 98 \, \text{m/s}^2 \)[/tex].
2. The gravitational acceleration at some height above the surface, [tex]\( g_{\text{height}} \)[/tex], is [tex]\( 2.5 \, \text{m/s}^2 \)[/tex].
3. The mass of the planet, [tex]\( m \)[/tex], is [tex]\( 6 \times 10^{24} \, \text{kg} \)[/tex].
4. The radius of the planet, [tex]\( R \)[/tex], is [tex]\( 6400 \, \text{km} \)[/tex] (which is [tex]\( 6400 \times 10^3 \, \text{m} \)[/tex]).
We need to find the height [tex]\( h \)[/tex] above the surface where the gravitational acceleration is [tex]\( 2.5 \, \text{m/s}^2 \)[/tex].
### Step-by-Step Solution:
1. Gravitational Force Formula:
The gravitational force experienced by an object at a distance [tex]\( r \)[/tex] from the center of a planet of mass [tex]\( M \)[/tex] is given by:
[tex]\[ g = \frac{G M}{r^2} \][/tex]
where [tex]\( G \)[/tex] is the gravitational constant, [tex]\( G \approx 6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex].
2. Gravitational Acceleration at Height [tex]\( h \)[/tex]:
At the height [tex]\( h \)[/tex] from the surface, the distance from the center of the planet is [tex]\( R + h \)[/tex].
[tex]\[ g_{\text{height}} = \frac{G M}{(R + h)^2} \][/tex]
3. Rearrange for [tex]\( R + h \)[/tex]:
We are given [tex]\( g_{\text{height}} = 2.5 \, \text{m/s}^2 \)[/tex].
[tex]\[ 2.5 = \frac{G \cdot 6 \times 10^{24}}{(6400 \times 10^3 + h)^2} \][/tex]
4. Solve for [tex]\( h \)[/tex]:
Rearrange the equation to solve for [tex]\( R + h \)[/tex]:
[tex]\[ 6400 \times 10^3 + h = \sqrt{\frac{G \cdot 6 \times 10^{24}}{2.5}} \][/tex]
Calculate the right-hand side:
[tex]\[ \sqrt{\frac{6.674 \times 10^{-11} \cdot 6 \times 10^{24}}{2.5}} \approx 12656065.739399428 \, \text{m} \][/tex]
5. Subtract the radius of the planet:
Subtract [tex]\( 6400 \times 10^3 \, \text{m} \)[/tex] to find the height [tex]\( h \)[/tex]:
[tex]\[ 12656065.739399428 \, \text{m} - 6400 \times 10^3 \, \text{m} = 6256065.739399428 \, \text{m} \][/tex]
So, the height [tex]\( h \)[/tex] above the planet's surface where the gravitational acceleration is [tex]\( 2.5 \, \text{m/s}^2 \)[/tex] is approximately [tex]\( 6256065.739399428 \, \text{m} \)[/tex].
We are given several key pieces of information:
1. The gravitational acceleration on the surface of a planet, [tex]\( g_{\text{surface}} \)[/tex], is [tex]\( 98 \, \text{m/s}^2 \)[/tex].
2. The gravitational acceleration at some height above the surface, [tex]\( g_{\text{height}} \)[/tex], is [tex]\( 2.5 \, \text{m/s}^2 \)[/tex].
3. The mass of the planet, [tex]\( m \)[/tex], is [tex]\( 6 \times 10^{24} \, \text{kg} \)[/tex].
4. The radius of the planet, [tex]\( R \)[/tex], is [tex]\( 6400 \, \text{km} \)[/tex] (which is [tex]\( 6400 \times 10^3 \, \text{m} \)[/tex]).
We need to find the height [tex]\( h \)[/tex] above the surface where the gravitational acceleration is [tex]\( 2.5 \, \text{m/s}^2 \)[/tex].
### Step-by-Step Solution:
1. Gravitational Force Formula:
The gravitational force experienced by an object at a distance [tex]\( r \)[/tex] from the center of a planet of mass [tex]\( M \)[/tex] is given by:
[tex]\[ g = \frac{G M}{r^2} \][/tex]
where [tex]\( G \)[/tex] is the gravitational constant, [tex]\( G \approx 6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex].
2. Gravitational Acceleration at Height [tex]\( h \)[/tex]:
At the height [tex]\( h \)[/tex] from the surface, the distance from the center of the planet is [tex]\( R + h \)[/tex].
[tex]\[ g_{\text{height}} = \frac{G M}{(R + h)^2} \][/tex]
3. Rearrange for [tex]\( R + h \)[/tex]:
We are given [tex]\( g_{\text{height}} = 2.5 \, \text{m/s}^2 \)[/tex].
[tex]\[ 2.5 = \frac{G \cdot 6 \times 10^{24}}{(6400 \times 10^3 + h)^2} \][/tex]
4. Solve for [tex]\( h \)[/tex]:
Rearrange the equation to solve for [tex]\( R + h \)[/tex]:
[tex]\[ 6400 \times 10^3 + h = \sqrt{\frac{G \cdot 6 \times 10^{24}}{2.5}} \][/tex]
Calculate the right-hand side:
[tex]\[ \sqrt{\frac{6.674 \times 10^{-11} \cdot 6 \times 10^{24}}{2.5}} \approx 12656065.739399428 \, \text{m} \][/tex]
5. Subtract the radius of the planet:
Subtract [tex]\( 6400 \times 10^3 \, \text{m} \)[/tex] to find the height [tex]\( h \)[/tex]:
[tex]\[ 12656065.739399428 \, \text{m} - 6400 \times 10^3 \, \text{m} = 6256065.739399428 \, \text{m} \][/tex]
So, the height [tex]\( h \)[/tex] above the planet's surface where the gravitational acceleration is [tex]\( 2.5 \, \text{m/s}^2 \)[/tex] is approximately [tex]\( 6256065.739399428 \, \text{m} \)[/tex].
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