Answer :
To determine the gravitational force between Mars and Phobos, we can use Newton's law of universal gravitation. The formula is given by:
[tex]\[ F = G \frac{m_1 \times m_2}{r^2} \][/tex]
Where:
- [tex]\( F \)[/tex] is the gravitational force.
- [tex]\( G \)[/tex] is the gravitational constant, [tex]\( 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex].
- [tex]\( m_1 \)[/tex] is the mass of Mars, [tex]\( 6.42 \times 10^{23} \, \text{kg} \)[/tex].
- [tex]\( m_2 \)[/tex] is the mass of Phobos, [tex]\( 1.06 \times 10^{16} \, \text{kg} \)[/tex].
- [tex]\( r \)[/tex] is the distance between Mars and Phobos, [tex]\( 9378 \, \text{km} \)[/tex] which needs to be converted to meters by multiplying by 1000 to get [tex]\( 9378000 \, \text{m} \)[/tex].
Now we plug the values into the formula:
[tex]\[ F = 6.67430 \times 10^{-11} \frac{(6.42 \times 10^{23}) (1.06 \times 10^{16})}{(9378000)^2} \][/tex]
First, calculate the product of the masses:
[tex]\[ m_1 \times m_2 = 6.42 \times 10^{23} \times 1.06 \times 10^{16} = 6.8052 \times 10^{39} \, \text{kg}^2 \][/tex]
Then, calculate the square of the distance:
[tex]\[ r^2 = (9378000)^2 = 8.7936484 \times 10^{13} \, \text{m}^2 \][/tex]
Now, substitute these values back in to find [tex]\( F \)[/tex]:
[tex]\[ F = 6.67430 \times 10^{-11} \frac{6.8052 \times 10^{39}}{8.7936484 \times 10^{13}} \][/tex]
Next, divide [tex]\( 6.8052 \times 10^{39} \, \text{kg}^2 \)[/tex] by [tex]\( 8.7936484 \times 10^{13} \, \text{m}^2 \)[/tex]:
[tex]\[ \frac{6.8052 \times 10^{39}}{8.7936484 \times 10^{13}} \approx 7.73844 \times 10^{25} \][/tex]
Then multiply by [tex]\( G \)[/tex]:
[tex]\[ F = 6.67430 \times 10^{-11} \times 7.73844 \times 10^{25} = 5.16 \times 10^{15} \, N \][/tex]
Hence, the gravitational force between Mars and Phobos is:
[tex]\[ 5.16 \times 10^{15} \, N \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{5.16 \times 10^{15} \, N} \][/tex]
[tex]\[ F = G \frac{m_1 \times m_2}{r^2} \][/tex]
Where:
- [tex]\( F \)[/tex] is the gravitational force.
- [tex]\( G \)[/tex] is the gravitational constant, [tex]\( 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex].
- [tex]\( m_1 \)[/tex] is the mass of Mars, [tex]\( 6.42 \times 10^{23} \, \text{kg} \)[/tex].
- [tex]\( m_2 \)[/tex] is the mass of Phobos, [tex]\( 1.06 \times 10^{16} \, \text{kg} \)[/tex].
- [tex]\( r \)[/tex] is the distance between Mars and Phobos, [tex]\( 9378 \, \text{km} \)[/tex] which needs to be converted to meters by multiplying by 1000 to get [tex]\( 9378000 \, \text{m} \)[/tex].
Now we plug the values into the formula:
[tex]\[ F = 6.67430 \times 10^{-11} \frac{(6.42 \times 10^{23}) (1.06 \times 10^{16})}{(9378000)^2} \][/tex]
First, calculate the product of the masses:
[tex]\[ m_1 \times m_2 = 6.42 \times 10^{23} \times 1.06 \times 10^{16} = 6.8052 \times 10^{39} \, \text{kg}^2 \][/tex]
Then, calculate the square of the distance:
[tex]\[ r^2 = (9378000)^2 = 8.7936484 \times 10^{13} \, \text{m}^2 \][/tex]
Now, substitute these values back in to find [tex]\( F \)[/tex]:
[tex]\[ F = 6.67430 \times 10^{-11} \frac{6.8052 \times 10^{39}}{8.7936484 \times 10^{13}} \][/tex]
Next, divide [tex]\( 6.8052 \times 10^{39} \, \text{kg}^2 \)[/tex] by [tex]\( 8.7936484 \times 10^{13} \, \text{m}^2 \)[/tex]:
[tex]\[ \frac{6.8052 \times 10^{39}}{8.7936484 \times 10^{13}} \approx 7.73844 \times 10^{25} \][/tex]
Then multiply by [tex]\( G \)[/tex]:
[tex]\[ F = 6.67430 \times 10^{-11} \times 7.73844 \times 10^{25} = 5.16 \times 10^{15} \, N \][/tex]
Hence, the gravitational force between Mars and Phobos is:
[tex]\[ 5.16 \times 10^{15} \, N \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{5.16 \times 10^{15} \, N} \][/tex]
