Satellite and Escape Velocity
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Direct answer
A satellite stays in orbit because gravity supplies exactly the centripetal pull it needs: GMm/r^2 = mv^2/r gives the orbital speed v_o = √(GM/r), about 7.9 km/s near the earth's surface, with period T = 2π√(r^3/GM) — the fact behind T^2 ∝ r^3. Escape demands v_e = √(2GM/R) = √(2gR), which for the earth is 11.2 km/s, exactly √2 times the surface orbital speed. The energy ledger is equally neat: KE = GMm/2r, PE = -GMm/r, total E = -GMm/2r, negative for a bound orbit, with the binding energy GMm/2r needed to free the satellite. A geostationary satellite combines these with T = 24 h to park over the equator at about 35,800 km height.
What you must remember
- Orbital speed: v_o = √(GM/r) = √(gR^2/r); independent of the satellite's own mass — a screwdriver and a station co-orbit.
- Escape velocity: v_e = √(2GM/R) = √(2gR) = 11.2 km/s for the earth; also independent of the projectile's mass and of the launch direction (energy is a scalar).
- The √2 link: v_e = √2 × v_o at the same radius — the single most quoted relation of this chapter.
- Orbital energies: KE = GMm/2r, PE = -GMm/r, E = -GMm/2r; adding energy makes E less negative and raises the orbit, it does not speed the satellite up — orbital speed actually falls as r grows.
- Geostationary conditions: T = 24 h, equatorial orbit, west-to-east sense of rotation; height about 35,800 km, radius about 42,000 km from the earth's centre.
- Weightlessness in orbit: not the absence of gravity but free fall — the satellite and astronaut accelerate together, so no normal force appears.
- Standard data: R = 6400 km, M = 6 × 10^24 kg, g = 9.8 m s^-2 — the triplet from which every NEET numerical starts.
From the surface to a 2R orbit, numerically
Place a 1000 kg satellite in a circular orbit of radius r = 2R (R = 6400 km). Its speed is v = √(GM/2R); using GM = gR^2 = 9.8 × (6.4 × 10^6)^2 ≈ 4.0 × 10^14, we get v = √(4.0 × 10^14/1.28 × 10^7) = √(3.14 × 10^7) ≈ 5.6 × 10^3 m s^-1, well below the 7.9 km/s surface value — higher orbits are slower orbits. The period follows as T = 2π√(r^3/GM) ≈ 2π√((1.28 × 10^7)^3/4 × 10^14) ≈ 14,300 s, roughly 4 hours. The energy ledger: E = -GMm/2r = -(4 × 10^14 × 1000)/(2 × 1.28 × 10^7) ≈ -1.56 × 10^10 J. That negative sign is information — to remove the satellite from orbit entirely you must supply +1.56 × 10^10 J, and to have parked it there from the ground you supplied more, because the launch also paid for the climb out of the earth's potential well.
How NEET frames the chapter
Three framings dominate. First, ratio questions: "a planet has twice the mass and half the radius of the earth — escape speed?" Since v_e ∝ √(M/R), the answer is √(2/0.5) = 2 times, 22.4 km/s — no G or constants needed. Second, the energy-sign trap: options include both -GMm/2r and -GMm/r, and candidates who skim pick the potential energy as the total; remember KE is half the magnitude of PE, making the total half of PE. Third, the escape-direction red herring: v_e is the same whether you fire the projectile up or at an angle, because escape is a work-energy statement, not a trajectory statement — though at exactly v_e the path is parabolic, and below it the path closes into an ellipse, a fact the options occasionally test. The moon having no atmosphere is why its escape velocity of about 2.4 km/s could not retain gases — a classic assertion-reason pairing.
Frequently asked questions
Does the orbital speed of a satellite depend on its mass?
No — the mass cancels between the gravitational force and the centripetal requirement, so v_o = √(GM/r) depends only on the central body and the orbital radius.
Why is escape velocity the same in every launch direction?
Because v_e = √(2GM/R) comes from equating kinetic energy to the depth of the gravitational potential well, and energy is independent of direction.
What is the total energy of a satellite in a circular orbit?
E = -GMm/2r — negative, showing a bound system, with KE exactly half the magnitude of the potential energy.
Why does a higher satellite move more slowly?
With v = √(GM/r), speed falls as radius rises; extra orbital energy makes the total less negative while the kinetic energy itself decreases.
What conditions make a satellite geostationary?
A 24-hour period, an orbit directly over the equator, and rotation in the same sense as the earth — satisfied at a height of about 35,800 km.