Gravity & g-Variation
Move a mass from a planet's core out into space and watch g and its weight change.
About this simulation
A cross-section of the planet paired with a live graph of g versus distance from the centre — rising linearly inside a uniform sphere, peaking at the surface, then falling as 1/r² outside. Moving the test mass with the distance slider reads off its weight at any depth or height, showing g peaks exactly at the surface.
How to use it
- Drag the Distance from centre slider from 0 to 3R and follow the dot along the g(r) graph; the planet diagram shows where the mass is.
- Read g, the weight W and the region at the top as you move, watch the mass cross the planet's edge in the diagram, and note that g is largest at r = R.
- Press the Try buttons 'Halfway to the core' and 'Two radii out' and compare the two g values with the surface value.
- Change Test mass and see that neither the graph nor the g value beside the dot moves: g does not depend on the mass being pulled, only the weight mg does.
- Raise Surface gravity, or press 'Jupiter-like surface', and watch every g value beside the dot scale up while the curve, drawn against g_s, keeps exactly the same shape.
What to notice
- Inside a uniform planet only the mass closer to the centre than you pulls you overall, so g falls in a straight line to zero at the centre.
- Outside the planet g follows the inverse-square law: at twice the radius it is a quarter of the surface value.
- The graph has a sharp corner at the surface, which is where g is greatest for a planet of uniform density.
- Weight W = mg changes from place to place, but the mass m of the test body does not.
The formulas
- g = g_s · r ⁄ R (inside, r ≤ R)
- Gravity at distance r from the centre of a uniform planet of radius R and surface gravity g_s.
- g = g_s · (R ⁄ r)² (outside, r ≥ R)
- Gravity outside the planet falls with the square of the distance r from its centre.
- W = mg
- Weight W of a body of mass m at a place where the acceleration due to gravity is g.
Questions
How does g change with depth below the Earth's surface?
Treating the Earth as a uniform sphere, g at depth d is g(1 − d/R), so it falls linearly and becomes zero at the centre. The shell of Earth above you exerts no net pull, leaving only the smaller inner sphere. In the simulation, move the mass from r = R towards the centre and the graph drops in a straight line.
How does g change with height above the Earth's surface?
At a height h, g = g_s R²/(R + h)², because the distance from the centre is now R + h. For heights much smaller than R this is close to g_s(1 − 2h/R). At h = R, twice the radius from the centre, g is one quarter of its surface value, as the 'Two radii out' preset shows.
What is the value of g at the centre of the Earth?
Zero. At the centre, the mass of the planet surrounds you equally on all sides, so the pulls cancel. Your weight would be zero there even though your mass is unchanged. Set Distance from centre to 0 in the simulation and the dot sits at the origin of the graph with g reading 0.00.
Where is the acceleration due to gravity maximum?
For a planet of uniform density, g is greatest exactly at the surface. Going down loses the pull of the outer layers, and going up increases the distance, so g decreases either way. The graph in the simulation peaks at r = R. The real Earth is denser at its core, so its g actually rises slightly for some depth before falling.
