Calculate the gravitational acceleration at the surface of the super-Earth GJ 1214B which has a mass of 5.7 MEarth and radius of 2.7REarth
Calculate the gravitational acceleration at the surface of the super-Earth GJ 1214B which has a mass of 5.7 MEarth and radius of 2.7REarth
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![**Example Problem: Calculating Gravitational Acceleration on a Super-Earth**
**Problem Statement:**
Calculate the gravitational acceleration at the surface of the super-Earth GJ 1214B, which has a mass of \(5.7 \, M_{\text{Earth}}\) and a radius of \(2.7 \, R_{\text{Earth}}\).
**Solution:**
To calculate gravitational acceleration, \(g\), at the surface of a planet, we use the formula:
\[ g = \frac{GM}{R^2} \]
where:
- \(G\) is the gravitational constant, \(6.674 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2\)
- \(M\) is the mass of the planet
- \(R\) is the radius of the planet
Given that:
- \(M_{\text{GJ 1214B}} = 5.7 \, M_{\text{Earth}}\)
- \(R_{\text{GJ 1214B}} = 2.7 \, R_{\text{Earth}}\)
We can substitute the values in terms of Earth's mass (\(M_{\text{Earth}}\)) and radius (\(R_{\text{Earth}}\)):
\[ g = \frac{G (5.7 \, M_{\text{Earth}})}{(2.7 \, R_{\text{Earth}})^2} \]
Simplifying this expression:
\[ g = \frac{5.7 \, GM_{\text{Earth}}}{2.7^2 \, R_{\text{Earth}}^2} \]
Knowing that gravitational acceleration on Earth \(g_{\text{Earth}}\) is
\[ g_{\text{Earth}} = \frac{GM_{\text{Earth}}}{R_{\text{Earth}}^2} \]
We can rewrite our expression as:
\[ g = \frac{5.7 \, g_{\text{Earth}}}{2.7^2} \]
Since \(g_{\text{Earth}} \approx 9.8 \, \text{m/s}^2\):
\[ g = \frac{5.7 \times 9.8}{2.7^2} \]
Calculating the numerical value:
\[ g = \frac{55.86}{7.29} \approx](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4fdfc26a-e6d8-4dfd-b2bf-654dccb4509d%2F1fedbfbe-eb13-44c4-95ab-a71a3199bde6%2Fp7w87ne_processed.png&w=3840&q=75)
Transcribed Image Text:**Example Problem: Calculating Gravitational Acceleration on a Super-Earth**
**Problem Statement:**
Calculate the gravitational acceleration at the surface of the super-Earth GJ 1214B, which has a mass of \(5.7 \, M_{\text{Earth}}\) and a radius of \(2.7 \, R_{\text{Earth}}\).
**Solution:**
To calculate gravitational acceleration, \(g\), at the surface of a planet, we use the formula:
\[ g = \frac{GM}{R^2} \]
where:
- \(G\) is the gravitational constant, \(6.674 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2\)
- \(M\) is the mass of the planet
- \(R\) is the radius of the planet
Given that:
- \(M_{\text{GJ 1214B}} = 5.7 \, M_{\text{Earth}}\)
- \(R_{\text{GJ 1214B}} = 2.7 \, R_{\text{Earth}}\)
We can substitute the values in terms of Earth's mass (\(M_{\text{Earth}}\)) and radius (\(R_{\text{Earth}}\)):
\[ g = \frac{G (5.7 \, M_{\text{Earth}})}{(2.7 \, R_{\text{Earth}})^2} \]
Simplifying this expression:
\[ g = \frac{5.7 \, GM_{\text{Earth}}}{2.7^2 \, R_{\text{Earth}}^2} \]
Knowing that gravitational acceleration on Earth \(g_{\text{Earth}}\) is
\[ g_{\text{Earth}} = \frac{GM_{\text{Earth}}}{R_{\text{Earth}}^2} \]
We can rewrite our expression as:
\[ g = \frac{5.7 \, g_{\text{Earth}}}{2.7^2} \]
Since \(g_{\text{Earth}} \approx 9.8 \, \text{m/s}^2\):
\[ g = \frac{5.7 \times 9.8}{2.7^2} \]
Calculating the numerical value:
\[ g = \frac{55.86}{7.29} \approx
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