Answer :
To find the mass of Mars using the given data about Phobos' orbit, we can apply Kepler's Third Law, which relates the orbital period of an object to its orbital radius and the mass of the central body (in this case, Mars).
Kepler's Third Law formula is given by:
[tex]\[ T^2 = \frac{4 \pi^2 r^3}{G M} \][/tex]
where:
- [tex]\( T \)[/tex] is the orbital period.
- [tex]\( r \)[/tex] is the orbital radius.
- [tex]\( G \)[/tex] is the gravitational constant, [tex]\( 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex].
- [tex]\( M \)[/tex] is the mass of Mars.
We need to solve for [tex]\( M \)[/tex]:
[tex]\[ M = \frac{4 \pi^2 r^3}{G T^2} \][/tex]
Step-by-Step Solution:
1. Identify the given values:
- Orbital period, [tex]\( T = 27553 \, \text{s} \)[/tex]
- Orbital radius, [tex]\( r = 9.378 \times 10^6 \, \text{m} \)[/tex]
2. Substitute the values into the formula:
[tex]\[ M = \frac{4 \pi^2 (9.378 \times 10^6 \, \text{m})^3}{(6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2})(27553 \, \text{s})^2} \][/tex]
3. Calculate the numerator [tex]\(4 \pi^2 r^3\)[/tex]:
[tex]\[ 4 \pi^2 (9.378 \times 10^6 \, \text{m})^3 \][/tex]
4. Calculate the denominator [tex]\( G T^2 \)[/tex]:
[tex]\[ (6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2})(27553 \, \text{s})^2 \][/tex]
5. Divide the two results to find [tex]\( M \)[/tex].
The resulting mass based on this calculation is:
[tex]\[ M \approx 6.426091629430167 \times 10^{23} \, \text{kg} \][/tex]
This is very close to one of the given options:
[tex]\[ 6.43 \times 10^{23} \, \text{kg} \][/tex]
Thus, the mass of Mars is [tex]\( 6.43 \times 10^{23} \, \text{kg} \)[/tex].
Kepler's Third Law formula is given by:
[tex]\[ T^2 = \frac{4 \pi^2 r^3}{G M} \][/tex]
where:
- [tex]\( T \)[/tex] is the orbital period.
- [tex]\( r \)[/tex] is the orbital radius.
- [tex]\( G \)[/tex] is the gravitational constant, [tex]\( 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex].
- [tex]\( M \)[/tex] is the mass of Mars.
We need to solve for [tex]\( M \)[/tex]:
[tex]\[ M = \frac{4 \pi^2 r^3}{G T^2} \][/tex]
Step-by-Step Solution:
1. Identify the given values:
- Orbital period, [tex]\( T = 27553 \, \text{s} \)[/tex]
- Orbital radius, [tex]\( r = 9.378 \times 10^6 \, \text{m} \)[/tex]
2. Substitute the values into the formula:
[tex]\[ M = \frac{4 \pi^2 (9.378 \times 10^6 \, \text{m})^3}{(6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2})(27553 \, \text{s})^2} \][/tex]
3. Calculate the numerator [tex]\(4 \pi^2 r^3\)[/tex]:
[tex]\[ 4 \pi^2 (9.378 \times 10^6 \, \text{m})^3 \][/tex]
4. Calculate the denominator [tex]\( G T^2 \)[/tex]:
[tex]\[ (6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2})(27553 \, \text{s})^2 \][/tex]
5. Divide the two results to find [tex]\( M \)[/tex].
The resulting mass based on this calculation is:
[tex]\[ M \approx 6.426091629430167 \times 10^{23} \, \text{kg} \][/tex]
This is very close to one of the given options:
[tex]\[ 6.43 \times 10^{23} \, \text{kg} \][/tex]
Thus, the mass of Mars is [tex]\( 6.43 \times 10^{23} \, \text{kg} \)[/tex].
