Questions: Mars has a mass of 6.42 × 10^23 kg. Phobos has a mass of 1.07 × 10^16 kg. The average distance from Mars's center to Phobol's center is 9.38 × 10^6 m. Calculate the gravitational force that Mars exerts on Phobos.

Solution Steps

We are given the following values:

• Mass of Mars, \( m_1 = 6.42 \times 10^{23} \, \text{kg} \)
• Mass of Phobos, \( m_2 = 1.07 \times 10^{16} \, \text{kg} \)
• Distance between the centers of Mars and Phobos, \( r = 9.38 \times 10^{6} \, \text{m} \)
• Gravitational constant, \( G = 6.67 \times 10^{-11} \, \frac{\text{m}^3}{\text{kg} \cdot \text{s}^2} \)

The formula for the gravitational force \( F_g \) between two masses is given by:

\[ F_{\mathrm{g}} = G \frac{m_{1} m_{2}}{r^{2}} \]

Substitute the given values into the formula:

\[ F_{\mathrm{g}} = \left(6.67 \times 10^{-11} \, \frac{\text{m}^3}{\text{kg} \cdot \text{s}^2}\right) \frac{(6.42 \times 10^{23} \, \text{kg})(1.07 \times 10^{16} \, \text{kg})}{(9.38 \times 10^{6} \, \text{m})^2} \]

First, calculate the product of the masses:

\[ m_1 \times m_2 = 6.42 \times 10^{23} \times 1.07 \times 10^{16} = 6.8694 \times 10^{39} \, \text{kg}^2 \]

Next, calculate the square of the distance:

\[ r^2 = (9.38 \times 10^{6})^2 = 8.8004 \times 10^{13} \, \text{m}^2 \]

Now, substitute these values back into the formula:

\[ F_{\mathrm{g}} = 6.67 \times 10^{-11} \times \frac{6.8694 \times 10^{39}}{8.8004 \times 10^{13}} \]

Calculate the force:

\[ F_{\mathrm{g}} = 6.67 \times 10^{-11} \times 7.8050 \times 10^{25} = 5.2044 \times 10^{15} \, \text{N} \]

Final Answer