Physics · Gravitation · NEET
No. When you equate gravitational force GMm/(R+h)^2 to centripetal force mv^2/(R+h), the mass m of the satellite appears on both sides and cancels out. So v = sqrt(GM/(R+h)) depends only on M (mass of earth) and the orbit radius (R+h), not on the satellite. A heavy satellite and a light satellite at the same height orbit at the same speed.
Gravity does. The earth's gravitational pull on the satellite points toward the earth's centre, and that is exactly the direction a centripetal force must point. So we write GMm/(R+h)^2 = mv^2/(R+h). Nothing else is needed — the satellite is in free fall, constantly falling toward earth but moving fast enough sideways to keep missing it.
Orbital velocity v_o = sqrt(GM/R) keeps a body in a closed circular orbit. Escape velocity v_e = sqrt(2GM/R) lets a body leave earth's gravity forever. Near the surface v_e = sqrt(2) * v_o, so escape velocity is about 1.414 times orbital velocity. For earth: v_o is about 7.9 km/s and v_e is about 11.2 km/s.
From v = sqrt(GM/(R+h)), a larger h makes the denominator (R+h) bigger, so v gets smaller. Higher satellites move slower. This is why low-orbit satellites zip around fast (about 90 minutes per orbit) while a geostationary satellite far out moves much slower and takes 24 hours.
Use GM = gR^2 (this comes from g = GM/R^2 at the surface). Substitute it into v = sqrt(GM/(R+h)) to get v = sqrt(gR^2/(R+h)) = R * sqrt(g/(R+h)). Right at the surface (h = 0) this simplifies to v = sqrt(gR).
A satellite of mass m orbits the earth (radius R) at height h. In terms of g0 (surface gravity), the total energy of the satellite is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
About 7.9 km/s (roughly 8 km/s). It comes from v = sqrt(gR) with g = 9.8 m/s^2 and R = 6.4 x 10^6 m, giving sqrt(9.8 x 6.4 x 10^6) which is about 7900 m/s.
v = sqrt(GM/(R+h)), where G is the gravitational constant, M is the mass of earth, R is earth's radius and h is the height of the orbit above the surface. Using GM = gR^2 it can also be written as v = R * sqrt(g/(R+h)), and at the surface v = sqrt(gR).
Here it is the speed (magnitude) needed for a circular orbit, so we treat it as a scalar value. The satellite's velocity vector is always tangent to the orbit (perpendicular to the radius), while gravity acts perpendicular to that velocity, toward the centre.
Orbital velocity is a high-yield gravitation topic. NEET regularly links it to satellite energy (KE, PE, total energy), time period, and escape velocity. The single derivation v = sqrt(GM/(R+h)) unlocks most satellite numericals, so mastering it saves time in the exam.
