Physics · Gravitation · NEET
g = GM/R² depends on mass, but mass is hidden inside the planet. For a uniform sphere M = ρ × (4/3)πR³. When you substitute this, mass cancels one power of R and you get g = (4/3)πGRρ. So density is just a cleaner way to write the same physics: it lets you compare two planets when you know their size (R) and how packed their matter is (ρ), without measuring mass directly.
Start with g = GM/R². Write the mass of a uniform sphere as M = ρ × V = ρ × (4/3)πR³. Substitute: g = G × [ρ × (4/3)πR³] / R². The R³/R² leaves one R, so g = (4/3)πGRρ. Rearranging for density gives ρ = 3g / (4πGR). Both forms carry equal marks in NEET — memorise the arrangement, not just the letters.
With density fixed, g = (4/3)πGRρ shows g ∝ R (directly proportional to radius). The 1/R² from g = GM/R² is misleading here because mass itself grows as R³ for a same-density planet. Net effect: g grows linearly with R. This is exactly what NEET tests in escape-velocity and 'same density, bigger planet' questions.
Rearrange g = (4/3)πGRρ to get ρ = 3g / (4πGR). Plugging g = 9.8 m/s², R = 6.4×10⁶ m and G = 6.67×10⁻¹¹ gives ρ ≈ 5.5×10³ kg/m³. This matches Earth's real mean density, which is why the formula is a favourite NEET single-line problem.
Not always — g depends on BOTH R and ρ through g = (4/3)πGRρ. A small very dense planet can have the same g as a large low-density one if the product Rρ is equal. In NEET you must compare the product R × ρ, never density alone.
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:
The escape velocity from the earth's surface is v. The escape velocity from the surface of another planet of radius four times that of earth and the same mean density is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
g = (4/3)πGRρ, where G is the gravitational constant, R is Earth's radius and ρ is its mean density. It is derived by putting M = ρ(4/3)πR³ into g = GM/R².
Using ρ = 3g/(4πGR) with g = 9.8 m/s², R = 6.4×10⁶ m, G = 6.67×10⁻¹¹, you get ρ ≈ 5.5×10³ kg/m³ (about 5500 kg/m³).
Yes, at fixed radius g ∝ ρ. And at fixed density g ∝ R. Both follow from g = (4/3)πGRρ.
NEET regularly asks to find Earth's density from g, or to scale g and escape velocity for a planet of same density but different radius. This single formula answers all of them quickly.
Yes. g = (4/3)πGRρ uses a uniform-density sphere. Real Earth is layered, so ρ here is the MEAN (average) density, which still gives the correct surface g.
