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![**Calculating Gravitational Acceleration on CoRoT 7b**
To calculate the gravitational acceleration at the surface of the super-Earth CoRoT 7b, which has a mass of 4.8 times the mass of Earth (\(4.8 \, M_{\text{Earth}}\)) and a radius of 1.7 times the Earth's radius (\(1.7 \, R_{\text{Earth}}\)), we use the formula for gravitational acceleration:
\[ g = \frac{GM}{R^2} \]
Here,
- \(G\) is the gravitational constant \( \approx 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)
- \(M\) is the mass of CoRoT 7b, which is \(4.8 \, M_{\text{Earth}}\)
- \(R\) is the radius of CoRoT 7b, which is \(1.7 \, R_{\text{Earth}}\)
The gravitational acceleration at the surface of CoRoT 7b can then be expressed as:
\[ g_{\text{CoRoT 7b}} = \frac{G \cdot (4.8 \, M_{\text{Earth}})}{(1.7 \, R_{\text{Earth}})^2} \]
This equation allows us to determine the surface gravity of super-Earth CoRoT 7b in comparison to Earth's gravity.](https://content.bartleby.com/qna-images/question/4fdfc26a-e6d8-4dfd-b2bf-654dccb4509d/94cb4a71-6abf-4f34-a3d2-3a42e14d70a6/thvpamx_processed.png)
Transcribed Image Text:**Calculating Gravitational Acceleration on CoRoT 7b**
To calculate the gravitational acceleration at the surface of the super-Earth CoRoT 7b, which has a mass of 4.8 times the mass of Earth (\(4.8 \, M_{\text{Earth}}\)) and a radius of 1.7 times the Earth's radius (\(1.7 \, R_{\text{Earth}}\)), we use the formula for gravitational acceleration:
\[ g = \frac{GM}{R^2} \]
Here,
- \(G\) is the gravitational constant \( \approx 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)
- \(M\) is the mass of CoRoT 7b, which is \(4.8 \, M_{\text{Earth}}\)
- \(R\) is the radius of CoRoT 7b, which is \(1.7 \, R_{\text{Earth}}\)
The gravitational acceleration at the surface of CoRoT 7b can then be expressed as:
\[ g_{\text{CoRoT 7b}} = \frac{G \cdot (4.8 \, M_{\text{Earth}})}{(1.7 \, R_{\text{Earth}})^2} \]
This equation allows us to determine the surface gravity of super-Earth CoRoT 7b in comparison to Earth's gravity.
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