Answer :
Sure! Let's go through the solution step-by-step to find the mass of the new planet in units of Earth masses.
1. Given values:
- Gravitational constant, \( G = 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)
- Radius of the Earth, \( R_{\text{Earth}} = 6.371 \times 10^6 \, \text{meters} \)
- Acceleration due to gravity on Earth, \( g_{\text{Earth}} = 9.81 \, \text{m/s}^2 \)
- Acceleration due to gravity on the new planet, \( g_x = 8.08 \, \text{m/s}^2 \)
2. Formula for gravitational acceleration:
The gravitational acceleration \( g \) at the surface of a planet is given by:
[tex]\[ g = \frac{G M}{R^2} \][/tex]
where \( M \) is the mass of the planet and \( R \) is the radius of the planet.
3. Calculate the mass of the new planet:
Rearrange the formula to solve for the mass \( M \):
[tex]\[ M = \frac{g R^2}{G} \][/tex]
For the new planet:
[tex]\[ M_{\text{new planet}} = \frac{g_x R_{\text{Earth}}^2}{G} \][/tex]
4. Calculate the mass of the Earth:
Using the same formula for Earth:
[tex]\[ M_{\text{Earth}} = \frac{g_{\text{Earth}} R_{\text{Earth}}^2}{G} \][/tex]
5. Calculate the mass of the new planet in units of Earth masses:
The mass of the new planet as a fraction of the Earth’s mass (denoted as \( f \)) is:
[tex]\[ f = \frac{M_{\text{new planet}}}{M_{\text{Earth}}} \][/tex]
6. Plug in the values and solve step-by-step:
[tex]\[ M_{\text{new planet}} = \frac{8.08 \times (6.371 \times 10^6)^2}{6.67430 \times 10^{-11}} \][/tex]
[tex]\[ M_{\text{Earth}} = \frac{9.81 \times (6.371 \times 10^6)^2}{6.67430 \times 10^{-11}} \][/tex]
Then:
[tex]\[ f = \frac{M_{\text{new planet}}}{M_{\text{Earth}}} \][/tex]
7. Result:
After performing these calculations, we find:
- Mass of the new planet: \( M_{\text{new planet}} \approx 4.91383814452452 \times 10^{24} \, \text{kg} \)
- Mass of the Earth: \( M_{\text{Earth}} \approx 5.965934677943755 \times 10^{24} \, \text{kg} \)
- Fraction of Earth’s mass: \( f \approx 0.8236493374108053 \)
Therefore, the mass of the new planet is approximately [tex]\( 0.824 \)[/tex] times the mass of the Earth.
1. Given values:
- Gravitational constant, \( G = 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)
- Radius of the Earth, \( R_{\text{Earth}} = 6.371 \times 10^6 \, \text{meters} \)
- Acceleration due to gravity on Earth, \( g_{\text{Earth}} = 9.81 \, \text{m/s}^2 \)
- Acceleration due to gravity on the new planet, \( g_x = 8.08 \, \text{m/s}^2 \)
2. Formula for gravitational acceleration:
The gravitational acceleration \( g \) at the surface of a planet is given by:
[tex]\[ g = \frac{G M}{R^2} \][/tex]
where \( M \) is the mass of the planet and \( R \) is the radius of the planet.
3. Calculate the mass of the new planet:
Rearrange the formula to solve for the mass \( M \):
[tex]\[ M = \frac{g R^2}{G} \][/tex]
For the new planet:
[tex]\[ M_{\text{new planet}} = \frac{g_x R_{\text{Earth}}^2}{G} \][/tex]
4. Calculate the mass of the Earth:
Using the same formula for Earth:
[tex]\[ M_{\text{Earth}} = \frac{g_{\text{Earth}} R_{\text{Earth}}^2}{G} \][/tex]
5. Calculate the mass of the new planet in units of Earth masses:
The mass of the new planet as a fraction of the Earth’s mass (denoted as \( f \)) is:
[tex]\[ f = \frac{M_{\text{new planet}}}{M_{\text{Earth}}} \][/tex]
6. Plug in the values and solve step-by-step:
[tex]\[ M_{\text{new planet}} = \frac{8.08 \times (6.371 \times 10^6)^2}{6.67430 \times 10^{-11}} \][/tex]
[tex]\[ M_{\text{Earth}} = \frac{9.81 \times (6.371 \times 10^6)^2}{6.67430 \times 10^{-11}} \][/tex]
Then:
[tex]\[ f = \frac{M_{\text{new planet}}}{M_{\text{Earth}}} \][/tex]
7. Result:
After performing these calculations, we find:
- Mass of the new planet: \( M_{\text{new planet}} \approx 4.91383814452452 \times 10^{24} \, \text{kg} \)
- Mass of the Earth: \( M_{\text{Earth}} \approx 5.965934677943755 \times 10^{24} \, \text{kg} \)
- Fraction of Earth’s mass: \( f \approx 0.8236493374108053 \)
Therefore, the mass of the new planet is approximately [tex]\( 0.824 \)[/tex] times the mass of the Earth.
