Which Has A Higher Gravitational Pull, A Stationary Or A Rotating Object? Why?

A rotating object has a very slightly greater gravitational pull than the same object at rest. Spinning adds rotational kinetic energy, and because energy has mass (E = mc²), that energy adds to the object’s total mass-energy and so to its gravity. The effect is far too tiny to notice for everyday objects. In Newtonian gravity, where pull depends only on mass, rotation makes no difference.

Over the centuries, various theories have been formulated to explain physical phenomena. One such phenomena is the attraction of objects towards the Earth. Many scientists and philosophers in the ancient world tried to explain this observation. In the 17^th century, Newton formulated the first explicit mathematical model for gravity in his book, Principles of Natural Philosophy, which was groundbreaking and launched a new epoch in natural sciences.

Newtonian gravity ruled the discussion until the 20^th century, when Einstein formulated his General Theory of Relativity.

General Relativity gives the most accurate picture of gravity to date, but is mathematically quite rigorous. For most everyday situations, Newtonian gravity works just fine without going very deep into challenging mathematics. So let’s start with Newton’s theory, because it explains why a spinning planet does not change how strongly its own gravity pulls. Here is the twist, though: Newton alone cannot actually settle the question in the title. For the honest answer we have to return to Einstein at the end. (If you want a deeper dive, you can read Einstein in his own words.)

Fundamentals Of Newtonian Gravity

Newtonian theory is based on observations of nature (empiricism). Gravity is a force that attracts two masses separated by some distance and always acts along the line joining the center of mass of both the objects. To be more precise, the attractive force between two objects is proportional to the product of the two masses and the reciprocal of distance squared. Mathematically, gravitational force, F:

where,

G = Universal Gravitational Constant ~ 6.674 × 10⁻¹¹ m³/kg/s²

m_a = mass of object A

m_b = mass of object B

r = distance between the two masses

Mechanics Of Rotating Objects

A mass rotating in a circular path experiences the following forces:

• An inertial force that acts along a tangential direction (at 90^o to the radius). If the mass were to be released from orbit, then it would move along the tangential path.
• A force directed towards the center of the circular path, along the radial line, is called centripetal force. This force is responsible for keeping the mass in orbit.
• A force directed away from the center of the circular path, along the radial line, is called centrifugal force. This is a fictitious (inertial) force that appears to push mass away from the center of orbit when you work in the rotating frame. Despite the common claim, it is not the Newtonian reaction to the centripetal force.

Mathematically, centripetal force, F_cp :

where,

m = mass of the object rotating in a circular trajectory

v = linear velocity of rotation

r = radius of circular path

ω = (v/r) = angular velocity (amount of rotation within a fixed time interval)

F_cp is a real force experienced by the object. A common mistake is to call the centrifugal force the Newton’s 3^rd law reaction to the centripetal force, but it is not, because an action and its reaction always act on two different bodies. The fictitious centrifugal force, F_cf, is instead an inertial (pseudo) force that only shows up when you do the bookkeeping from inside the rotating frame itself. It is equal in magnitude, but opposite in direction to F_cp .

Therefore,

Here, the negative sign (-) signifies the opposite direction to F_cp .

Dynamics Of A Two-Body System

In physics, whenever two or more forces act on a particle simultaneously, the resultant force is the vector sum of the individual forces. A vector sum is a mathematical operation similar (but not identical) to simple algebraic addition, but also takes into account the direction of the forces.

For easier calculations, let’s assume the presence of a perfect sphere, having mass M and radius R. The sphere can be modeled as a collection of infinitesimally thin (extremely thin) circular disks placed on top of each other. The radius of the central disk (equatorial line) is maximum, i.e., R, and the radius of the poles is zero.

Before diving in, one clarification matters: the calculation below tracks the net force felt by a small test mass m sitting on (or orbiting) the big sphere. That is its apparent weight, which is what you would read on a bathroom scale. It is not the gravitational pull the sphere exerts on the outside universe, a distinction we return to at the end. Consider the following cases:

CASE 1: The sphere does not rotate

The gravitational force experienced by a small object of mass m, at a distance r from M is:

Here, M>>>m and r>>>R (this approximation is analogous to satellites orbiting the Earth).

Since the sphere is not rotating, the centrifugal force, F_cf, is zero. Net force on m:

CASE 2: The sphere starts rotating

Now, assume that the sphere of mass M and radius R starts rotating, with its axis of rotation along the +z direction. Gravitational force is:

Centripetal and centrifugal forces start acting on the surface of the sphere. F_cf is maximum at the equator (−Mω²R) and minimum at the poles (approaching zero). The magnitude of centrifugal force, F_cf at the equator and poles is given by:

Therefore, the net force on m due to gravitational attraction from M and the rotation of M about the +z axis is:

The Answer (As Far As Newton Is Concerned)

In the rotating frame, the centrifugal force shaves a little off the net force felt by m at the equatorial plane. That net force grows as you move toward the poles, where it reaches its maximum and equals the full gravitational force. The shortfall is tiny for a slow rotator like Earth. Measured at sea level, the acceleration due to gravity, g, is about 9.798 m/s² at the equator and 9.863 m/s² at the poles. (Two effects combine here: the centrifugal force and the fact that the equator, sitting on Earth’s bulge, is farther from the center than the poles are.)

But notice what this actually tells us. The centrifugal term changes the apparent weight of an object resting on the spinning sphere. It does not change the mass M in F = G Mm/r², so the gravitational pull the sphere exerts on the rest of the universe is exactly the same whether it spins or not. In short, within Newtonian gravity, rotation makes no difference to an object’s gravitational pull at all, because that pull depends only on its mass. So Newton, strictly speaking, says “no difference,” and to do better we have to call in Einstein.

So Does Spinning Change Gravity At All? The Relativistic Answer

Here is where it gets interesting. In general relativity, gravity is sourced not by mass alone but by all forms of energy and momentum (bundled together in something called the stress-energy tensor). And a spinning object carries extra energy that a stationary one does not: its rotational kinetic energy.

Einstein’s most famous equation, E = mc², tells us that energy has mass. So that rotational energy E_rot behaves like an added sliver of mass, E_rot/c², bumping up the object’s total mass-energy. More mass-energy means a marginally stronger gravitational field. So a rotating object really does have a very slightly greater gravitational pull than the same object sitting still.

How slight? Absurdly slight. Because we divide by c² (roughly 9 × 10¹⁶ in SI units), the spin of an everyday object, a flywheel, a planet, even a star, adds a vanishingly small fraction to its mass. For Earth, the rotational energy works out to something like 10^-13 of its rest mass-energy. No instrument could weigh the difference; it is a matter of principle, not of practice. But the principle is the real answer to the title’s question, and it points the opposite way from what the simple centrifugal picture might suggest.

There is a second, completely separate relativistic effect worth knowing about. A rotating mass does not just sit in spacetime; it drags spacetime around with it, a bit like a spinning ball stirring honey. This is called frame-dragging (or the Lense-Thirring effect), and it is described exactly by the Kerr metric, the relativistic solution for a spinning body. Frame-dragging is not the same thing as the energy effect above and it does not make the pull stronger; it twists the geometry around the spin axis. It is real, though: NASA’s Gravity Probe B mission, launched in 2004, used ultra-precise gyroscopes in orbit and in 2011 confirmed Earth’s frame-dragging at about −37.2 milliarcseconds per year, close to the −39.2 mas/yr that relativity predicts.

So the full answer has two layers. Ask Newton, and rotation changes nothing about gravitational pull. Ask Einstein, and a spinning object has an immeasurably tiny bit more pull (thanks to its rotational energy), plus it warps the spacetime around it in a way a stationary object never could.