If you leave the radius of a planet the same but double its mass, how does the gravitational pull of the planet on a person (equivalent to the person's weight) change? If you leave the mass the same, but double the radius how does the gravitational pull of the planet change?
thanks
Force=GMe*m/re^2
double M, double F
double re, F is 1/4 original
huh? we haven't learned any of these formulas..
SInce gravity gets weaker with distance, if you doubled the radius wouldn't you feel your weight less and less?
F = G*(M*m/r^2)
he's just saying that (according to a formula you've maybe never seen)
if you double the planet's mass (M in the formula) you will double the force due to gravity, because the relationship is linear
if you double the planet's radius, you will reduce the force due to gravity by a factor of 4, because r^2 is in the denominator
To understand how the gravitational pull of a planet on a person changes when the planet's mass or radius is altered, we need to consider Newton's law of universal gravitation:
F = (G * m1 * m2) / r^2
where F is the gravitational force between two objects, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers.
Let's examine the two scenarios you mentioned:
1. If you leave the radius of the planet the same but double its mass:
In this case, if we denote the original mass as M1 and the double mass as M2, we can compare the gravitational forces by setting up the equations:
F1 = (G * m1 * M1) / r^2
F2 = (G * m1 * M2) / r^2
Dividing the two equations, we get:
F2/F1 = M2/M1
Thus, the gravitational force of the planet on a person (equivalent to their weight) will also double. It increases directly proportional to the increase in mass.
2. If you leave the mass the same but double the radius:
Similarly, if we denote the original radius as r1 and the double radius as r2, we can set up the equations:
F1 = (G * m1 * M1) / r1^2
F2 = (G * m1 * M1) / r2^2
Dividing the two equations, we get:
F2/F1 = r1^2/r2^2
Simplifying, we find:
F2/F1 = 1/(r2/r1)^2
As a result, the gravitational force decreases proportionally to the square of the ratio of the initial radius to the expanded radius. In this case, it will be one-fourth (1/[(r2/r1)^2]) of the original force.
Therefore, if you double the mass while keeping the radius constant, the gravitational pull on a person will double. If you double the radius while keeping the mass constant, the gravitational pull will become one-fourth of its initial value.