Question

A 4525kg rocket orbits the Earth with a velocity of 7825m/s. (RE=6.371x106m, ME=5.96x1024kg). What is the radius of the rockets orbit.

Answer 1

Approximately
`6.49 * 10^(6)\; {\rm m}`.

Step-by-step explanation:

Look up the Gravitational Constant:

`G \approx 6.67 * 10^(-11)\; {\rm m^(3)\, kg^(-1)\, s^(-2)}`.

Assume that the net force
`F_{\text{net}}` on this rocket is equal to the gravitational attraction from the Earth (i.e., there is no other force on this rocket.) This force would be equal to:

`\displaystyle F_{\text{net}} = (G\, M\, m)/(r^(2))`,

Where:

• 
  `G \approx 6.67 * 10^(-11)\; {\rm m^(3)\, kg^(-1)\, s^(-2)}` is the gravitational field strength,
• 
  `M \approx 5.96 * 10^(24)\; {\rm kg}` (as given) is the mass of the Earth,
• 
  `m` is the mass of this rocket, and
• 
  `r` is the orbital radius that needs to be found.

Since the rocket is in a circular orbit of radius
`r` with a tangential speed of
`v`, acceleration would be equal to:

`\displaystyle a = (v^(2))/(r)`.

The net force on this rocket would be equal to:

`\displaystyle F_{\text{net}} = m\, a = (m\, v^(2))/(r)`.

Equate the two expressions for the net force on the rocket (from the gravitational force and from the centripetal motion) to obtain:

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

Simplify and solve this equation for orbital radius
`r`:

`\begin{aligned}(G\, M)/(r^(2)) = (v^(2))/(r)\end{aligned}`.

`\begin{aligned}r &= (G\, M)/(v^(2)) \\ &\approx \frac{(6.67 * 10^(-11)\; {\rm m^(3)\, kg^(-1)\, s^(-2)})\, (5.69 * 10^(24)\; {\rm kg})}{(7825\; {\rm m\cdot s^(-1)})^(2)} \\ &\approx 6.49 * 10^(6)\; {\rm m}\end{aligned}`.