Questions: Orbital Mechanics: Circular and Elliptical Orbits

Question 1

A satellite is moved from a circular orbit of radius r to a new circular orbit of radius 4r. By what factor does its orbital speed change?

• A. It increases by a factor of 2
• B. It decreases by a factor of 2
• C. It decreases by a factor of 4
• D. It stays the same

Answer: (B)

Orbital speed v = √(GM/r), so v ∝ 1/√r. When r increases by a factor of 4, v changes by a factor of 1/√4 = 1/2. The satellite in the larger orbit moves half as fast. This is counterintuitive: boosting a satellite to a higher orbit requires adding energy, yet the satellite ends up moving more slowly.

Question 2

Astronauts aboard the International Space Station experience weightlessness because there is no significant gravity at that altitude.

• T. True
• F. False

Answer: False

At ISS altitude (~400 km), Earth's gravitational field is roughly 90% as strong as on the surface. Astronauts feel weightless because the station and everything in it is in free fall — all objects accelerate toward Earth at the same rate. There is no normal force from the floor because the floor is also falling. Weightlessness is the sensation of free fall, not the absence of gravity.

Question 3

Why is the total mechanical energy of a circular orbit negative?

Model answer: A negative total energy indicates a bound orbit: the satellite lacks the kinetic energy to escape the gravitational potential well. E = KE + PE = GMm/(2r) − GMm/r = −GMm/(2r). The magnitude of potential energy exceeds kinetic energy, so the total is negative.

The sign of total mechanical energy distinguishes bound from unbound trajectories. E < 0 means the orbit is bound (circle or ellipse). E = 0 corresponds to an escape trajectory (parabola). E > 0 means the object escapes with kinetic energy to spare (hyperbola). This follows from how gravitational potential energy (negative by convention) relates to kinetic energy in orbit.