Orbital motion

Satellites are not weightless because "gravity is zero in space". Gravity weakens with distance but is never zero (at the ISS, $g \approx 8.7$ N/kg) and IS the centripetal force holding the orbit. Astronauts feel weightless only because they and the craft free-fall together, so there is no contact force — the apparent weight, not gravity, is zero.

A geosynchronous orbit has a 24 h period but any inclination (it traces a figure-8, not fixed). A geostationary orbit is the special equatorial case that stays above one point. The period is exactly 24 h — not "same as the Moon" (27.3 days) or "1 year". A full definition needs three elements: 24 h period, equatorial plane, same direction as Earth's spin.

In $v = 2\pi r/T$, convert before substituting: radius in metres (not km), period in seconds (not hours or minutes). A period left in hours is out by 3600; a radius in km by 1000. These slips lose the final answer mark even when the method is right — and the answer should be in m/s.

Intuition says "lower = slower", but orbits are the opposite. $v = \sqrt{GM/r}$: a smaller $r$ gives a larger $v$ and a shorter period. The ISS (low) orbits in ~90 min; the Moon (far) takes 27.3 days. So a satellite moved to a lower orbit speeds up, it does not slow down.

Orbital radius $r$ is measured from the centre of the Earth, so $r = \text{altitude} + R_\text{Earth}$ with $R_\text{Earth} \approx 6.4 \times 10^6$ m. Substituting the altitude (measured from the surface) makes $r$ about 6400 km too small. Always convert altitude to orbital radius before using $v = 2\pi r/T$.

On an orbit diagram the velocity arrow is tangential (along the orbit) and the gravitational/centripetal force arrow points inward to the centre. Drawing the velocity towards Earth, or the force along the direction of motion, is penalised. Label the inward force as gravity acting as the centripetal force.