Question

A satellite is in a circular orbit around Mars, which has a mass M = 6.40 × 1023 kg and radius R = 3.40 ×106 m.

a) Derive an expression for the orbital speed assuming that the orbital radius is R.

b) Derive an expression for the orbital period assuming that the orbital radius is R.

c) The orbital period of the satellite can be synchronized with Mars’ rotation, whose period is 24.6 hours. Determine the required orbital radius for this to occur.

d) Determine the escape speed from the surface of Mars.

Answer 1

a) The orbital speed of a satellite with a orbital radius R (in meters) will have an orbital speed of approximately
`\displaystyle \sqrt(4.27 * 10^(13))/(R)\; \rm m \cdot s^(-1)`.

b) Again, if the orbital radius R is in meters, the orbital period of the satellite would be approximately
`\displaystyle 9.62 * 10^(-7)\, R^(3/2)\; \rm s`.

c) The orbital radius required would be approximately
`\rm 2.04 * 10^7\; m`.

d) The escape velocity from the surface of that planet would be approximately
`\rm 5.01* 10^3\; m \cdot s^(-1)`.

Step-by-step explanation:

a)

Since the orbit of this satellite is circular, it is undergoing a centripetal motion. The planet's gravitational attraction on the satellite would supply this centripetal force.

The magnitude of gravity between two point or spherical mass is equal to:

`\displaystyle (G \cdot M \cdot m)/(r^(2))`,

where

• 
  `G` is the constant of universal gravitation.
• 
  `M` is the mass of the first mass. (In this case, let
  `M` be the mass of the planet.)
• 
  `m` is the mass of the second mass. (In this case, let
  `m` be the mass of the satellite.)
• 
  `r` is the distance between the center of mass of these two objects.

On the other hand, the net force on an object in a centripetal motion should be:

`\displaystyle (m \cdot v^(2))/(r)`,

where

• 
  `m` is the mass of the object (in this case, that's the mass of the satellite.)
• 
  `v` is the orbital speed of the satellite.
• 
  `r` is the radius of the circular orbit.

Assume that gravitational force is the only force on the satellite. The net force should be equal to the planet's gravitational attraction on the satellite. Equate the two expressions and solve for
`v`:

`\displaystyle (G \cdot M \cdot m)/(r^(2)) = (m \cdot v^(2))/(r)`.

`\displaystyle v^2 = (G \cdot M)/(r)`.

`\displaystyle v = \sqrt{(G \cdot M)/(r)}`.

Take
`G \approx 6.67 * \rm 10^(-11) \; m^3 \cdot kg^(-1) \cdot s^(-2)`, Simplify the expression
`v`:

`\begin{aligned} v &= \sqrt{(G \cdot M)/(r)} \cr &= \sqrt{(6.67 * \rm 10^(-11) * 6.40 * 10^(23))/(r)} \cr &\approx \sqrt{(4.27 * 10^(13))/(r)} \; \rm m \cdot s^(-1) \end{aligned}`.

b)

Since the orbit is a circle of radius
`R`, the distance traveled in one period would be equal to the circumference of that circle,
`2 \pi R`.

Divide distance with speed to find the time required.

`\begin{aligned} t &= (s)/(v) \cr &= 2 \pi R}\left/\sqrt{(G \cdot M)/(R)} \; \rm m \cdot s^(-1)\right. \cr &= (2\pi R^(3/2))/(√(G \cdot M)) \cr &\approx  9.62 * 10^(-7)\, R^(3/2)\; \rm s\end{aligned}`.

c)

Convert
`24.6\; \rm \text{hours}` to seconds:

`24.6 * 3600 = 88560\; \rm s`

Solve the equation for
`R`:

`9.62 * 10^(-7)\, R^(3/2)= 88560`.

`R \approx 2.04 * 10^7\; \rm m`.

d)

If an object is at its escape speed, its kinetic energy (KE) plus its gravitational potential energy (GPE) should be equal to zero.

`\displaystyle \text{GPE} = -(G \cdot M \cdot m)/(r)` (Note the minus sign in front of the fraction. GPE should always be negative or zero.)

`\displaystyle \text{KE} = (1)/(2) \, m \cdot v^(2)`.

Solve for
`v`. The value of
`m` shouldn't matter, for it would be eliminated from both sides of the equation.

`\displaystyle -(G \cdot M \cdot m)/(r) + (1)/(2) \, m \cdot v^(2)= 0`.

`\displaystyle v = \sqrt{(2\, G \cdot M)/(R)} \approx 5.01* 10^(3)\; \rm m\cdot s^(-1)`.