Q: An arbitrary distance separates two objects of equal mass. If the mass of each object is doubled, and the distance between the two objects is tripled, how will the force of attraction between the two objects change?
A: This question regards the gravitational force of attraction that exists between two objects with well-defined masses ( unlike a massless photon that is also influenced by gravitational forces ). An object’s velocity ( v ) undergoes a constant increase or decrease under the influence of a force equal to F = ( m )( Δv/Δt ) = ma. The SI unit of force is the Newton ( N ), m = mass in kilograms ( kg ), and a = acceleration in meters per second squared ( m/s^2 ).
The gravitational force of attraction between two masses is quantified by F = G(m1 )( m2 )/r^2, where G ( 6.67408 x 10^-11 m^3 kg^-1 s^-2 ) is the universal gravitational constant. Near the earth’s surface, the radial distance ( r ) between a falling object and the earth’s surface is approximately equal to the earth’s radius. Since the mass ( M ) of the earth is constant, G( M )/r^2 reduces to a gravitational constant of acceleration ( g ) = ( 9.8 m/s^2 ), and F = ( m2 )( g ). If the mass of a particle were doubled, would its acceleration towards the earth double? Well, the answer is no, and here’s why.
Newton’s Third Law of Motion states that forces occur in pairs. A mass that falls near the earth’s surface has a LARGE force exerted upon it by the earth; however, the particle exerts an equal and opposite force upon the earth! How is this possible? Well, let’s see what mathematics tells us.
If F1 = F2, then ( M1 )( a1 ) = ( m2 )( g ), where M1 is the mass of the earth, and m2 is the mass of a falling object. Notice that M1 is much larger than m2. Thus, when we solve for a1, we have a1 = ( m2 / M1 )( g ). Thus, the rate at which the earth accelerates towards falling bodies is negligible.
Now, let’s examine a scenario in which two particles fall towards earth ( neglecting wind resistance ). Each particle has a mass of m1, and they each fall towards earth under the influence of a force F = ( m1 )( g ). If these masses combined, we’d have 2F = ( 2m1 )( g ). Therefore, the force exerted by the earth upon our new mass of 2m1 has doubled; however, the force that our new mass exerts upon the earth would double as well. Thus, if F1 = F2, then 2F1 = 2F2, and ( 2F2 / 2m1 ) still equals g! Since mass is a measure of inertia ( I ), larger forces must act upon larger masses near the earth’s surface to make them accelerate with a magnitude of 9.8 m/s^2. As a consequence, it is more difficult to bring larger falling masses to rest.
Furthermore, a quantity of work ( W ) is required to lift an object against the earth’s gravitational force of attraction. Work ( W ) = Fd, where F = Force and d = distance ( height ) in meters ( m ). The unit of energy needed to change the position of a mass over a given distance is measured in SI units of Joules ( J ). If an object of mass m1 falls from some height ( h ), it’s initial potential energy ( PE ) = mgh. After falling through h, an object’s initial potential energy is converted to an ” energy of motion ” referred to as kinetic energy ( KE ) = 1/2 mv^2, where m = mass in kilograms ( kg ), and v = the final velocity of the object in meters per second squared ( m/s )^2.
Doubling the mass of the aforementioned object would double its initial potential energy. As a consequence, the mass will have twice the kinetic energy as the original mass. If we set 2PE = 2KE and solve for final velocity, the velocity will be no different than the velocity derived from PE = KE. Differing masses must accelerate at g = 9.8 m/s^2, because differing masses accelerating at differing rates-of-change would violate laws regarding the Conservation of Energy.
If the mass of two objects are doubled, and the distance between the two objects is tripled, then F = G( 2m1 )( 2m2 )/( 3 m )^2 .
Thus, the final force = ( 4/9 )( F ).
RENAISSANCE PHYSICS 