Physics · Gravitation · NEET
Start with T = 2*pi*sqrt(R^3 / (G*M)). Replace the mass using density: M = (4/3)*pi*R^3*d. Substituting gives T = 2*pi*sqrt(R^3 / (G*(4/3)*pi*R^3*d)). The R^3 in the top and bottom cancel completely, leaving T = sqrt(3*pi / (G*d)). So for a satellite skimming the surface, only the mean density d matters, not the planet's size.
For an orbit radius equal to R (just above the surface): T = sqrt(3*pi / (G*d)), which is the same as T^2 = 3*pi / (G*d). Here G is the gravitational constant and d is the mean density of the planet. This is the form NEET tests directly.
No. This clean result only holds when the orbit radius equals the planet's radius R (satellite just above the surface). For a satellite at height h, the orbit radius is r = R + h and R does NOT cancel, so you must use the full T = 2*pi*sqrt(r^3 / (G*M)) and the period grows with height.
Kepler's third law says T^2 is proportional to r^3. For a near-surface orbit r = R, and because M itself contains R^3 through density, the R^3 factors cancel and the proportionality collapses to a constant that depends only on d. So this is Kepler's law specialised to the surface orbit.
Shorter. Since T = sqrt(3*pi / (G*d)), a larger density d makes the denominator bigger, so T becomes smaller. A denser planet pulls harder for the same size, so the skimming satellite must move faster and completes an orbit in less time.
A satellite orbits just above the earth's surface with period T. If d is the mean density of the earth and G the gravitational constant, the quantity 3*pi/(G*d) represents:
If R is the radius of the earth and g the acceleration due to gravity on its surface, the mean density of the earth is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
About 84 to 85 minutes. Using T = sqrt(3*pi/(G*d)) with Earth's mean density d = 5500 kg/m^3 and G = 6.67e-11, you get roughly 5060 seconds, close to 84 minutes. This is the shortest possible period for an Earth satellite.
Yes. It is a standard derived result from the Gravitation chapter and has been asked directly in NEET 2023. It combines the satellite period formula with the density expression for mass, both from NCERT Class 11 Physics.
Mass and radius are linked through density: M = (4/3)*pi*R^3*d. When both appear in the period formula for a surface orbit, the R^3 terms cancel, and mass gets absorbed into density. What is left is a single dependence on d.
A geostationary satellite has a period of 24 hours at a height of about 36000 km, so its orbit radius is much larger than R and R does not cancel. The near-surface case gives the minimum period of about 84 minutes and depends only on density.
