Mars has a mass of about \(6.42 \times 10^{23} \mathrm{kg} .\) The length of a day on Mars is \(24 \mathrm{h}\) and 37 min, a little longer than the length of a day on Earth. Your task is to put a satellite into a circular orbit around Mars so that it stays above one spot on the surface, orbiting Mars once each Mars day. At what distance from the center of the planet should you place the satellite?

The gravitational force between two objects is given by Newton's law of gravitation. The formula for this force is: \(F = G \frac{m_1 m_2}{r^2}\) where \(F\) is the force, \(G\) is the gravitational constant \(6.674 \times 10^{-11}\,\text{N}\,\text{m}^2\,\text{kg}^{-2}\), \(m_1\) and \(m_2\) are the masses of the two objects (in this case Mars and the satellite), and \(r\) is the distance between their centers.

The centripetal force acting on the satellite is given by the formula: \(F_c = m\,a = m\,\frac{v^2}{r}\) Where \(m\) is the mass of the satellite, \(a\) is the centripetal acceleration, and \(v\) is the orbital velocity.

The gravitational force is the only force acting on the satellite, so it must be equal to the centripetal force: \(F = F_c\) \(G \frac{m_1 m_2}{r^2} = m\,\frac{v^2}{r}\)

To find the orbital velocity, we use the formula for the circumference of the orbit divided by the orbital period: \(v = \frac{2 \pi r}{T}\) Where \(T\) is the orbital period, which is equal to the length of a Mars day (24 hours and 37 minutes). We want the orbital period to be the same as a day on Mars, so we need to convert the hours and minutes into seconds: \(T = (24 * 3600) + (37 * 60) \approx 88740\,s\) Now we substitute this expression for \(v\) back into the equation from Step 3: \(G \frac{m_1 m_2}{r^2} = m\,\frac{(2 \pi r / T)^2}{r}\)

Now we can simplify the equation and solve for \(r\): \(G \frac{m_1 m_2}{r^2} = m\,\frac{(4 \pi^2 r^2) / T^2}{r}\) Notice that the mass of the satellite \(m\) cancels out, so we're left with: \(G \,m_1 = \frac{4 \pi^2 r^3}{T^2 r}\) Now we can solve for the distance \(r\): \(r^3=\frac{G\, m_1\, T^2}{4\,\pi^2}\) \(r=\sqrt[3]{\frac{G\, m_1\, T^2}{4\,\pi^2}}\) Plugging in the values for Mars mass, \(G\), and \(T\): \(r=\sqrt[3]{\frac{6.674 \times 10^{-11}\,\text{N}\,\text{m}^2\,\text{kg}^{-2} \times 6.42 \times 10^{23}\,\text{kg} \times (88740\,s)^2}{4\,\pi^2}}\) Evaluating this expression gives: \(r \approx 20430000\,\text{m}\) So the satellite should be placed approximately 20,430,000 meters from the center of Mars to have a circular orbit with a period equal to one Mars day.