What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?

Wait a moment while we are looking for your question.

.

Engineering

Physics

Gravitational Force and Field

Gravitational Acceleration

Artificial Satellites

Question

What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?

A

$\frac{5 GmM}{6 R}$

B

$\frac{2 GmM}{3 R}$

C

$\frac{GmM}{2 R}$

D

$\frac{GmM}{3 R}$

Required energy $- \frac{GMm}{R} = \frac{- GMm}{6 R}$

To find the minimum energy required to launch a satellite from the surface of a planet into a circular orbit at an altitude of 2R, we need to calculate the total energy change. The minimum energy corresponds to the difference between the satellite's total mechanical energy in its final orbit and its total mechanical energy on the planet's surface.

Step 1: Determine the total energy on the surface of the planet.
On the surface, the satellite is at rest relative to the planet. Its total energy (E_surface) is only the gravitational potential energy because kinetic energy is zero (for minimum launch).
Gravitational potential energy at the surface (distance R from center) is given by:
$U_{surface} = - \frac{G m M}{R}$
So, E_surface = $- \frac{G m M}{R}$

Step 2: Determine the total energy in the circular orbit at altitude 2R.
The orbital radius is R + 2R = 3R.
For a circular orbit, the centripetal force is provided by gravitational force:
$\frac{G m M}{{(3 R)}^{2}} = \frac{m v^{2}}{3 R}$
Solving for kinetic energy (K_orbit):
$\frac{1}{2} m v^{2} = \frac{1}{2} \frac{G m M}{3 R} = \frac{G m M}{6 R}$
Gravitational potential energy at orbit (U_orbit) is:
$U_{orbit} = - \frac{G m M}{3 R}$
Total energy in orbit (E_orbit) = K_orbit + U_orbit = $\frac{G m M}{6 R} - \frac{G m M}{3 R} = - \frac{G m M}{6 R}$

Step 3: Calculate the minimum energy required for launch.
This is the energy that must be provided to change the satellite's total energy from E_surface to E_orbit.
Minimum energy required = E_orbit - E_surface = $(- \frac{G m M}{6 R}) - (- \frac{G m M}{R}) = - \frac{G m M}{6 R} + \frac{G m M}{R} = \frac{G m M}{R} - \frac{G m M}{6 R} = \frac{5 G m M}{6 R}$

Final Answer: The minimum energy required is $\frac{5 G m M}{6 R}$

Related Topics and Formulae

Gravitational Potential Energy: The potential energy associated with the gravitational force between two masses. For a point mass or outside a spherical body, it is given by $U = - \frac{G M m}{r}$ , where r is the distance between centers.

Orbital Mechanics: For a satellite in a circular orbit, the centripetal force equals the gravitational force: $\frac{G M m}{r^{2}} = \frac{m v^{2}}{r}$ . This leads to the orbital velocity $v = \sqrt{\frac{G M}{r}}$ .

Total Energy in Orbit: The total mechanical energy (kinetic + potential) for a satellite in a circular orbit is always negative and given by $E = - \frac{G M m}{2 r}$ . This indicates a bound system.