# Question related to accleration due to gravity [closed]

What effect will be seen on the value of accleration due to gravity if radius of the earth is decreased keeping the mass constant?

If you stayed at the same radius while the earth shrinks then nothing will change other than you'll start to fall. If you stand on the surface of the new smaller radius then you will feel the increased gravitation.

The strength of the gravitational field depends on how far you are from the center of the mass, assuming the mass density is radially dependent or constant. It also depends on the total mass inside your position: $$g=\frac{GM_E}{r^2},$$ where $r$ is the distance from the center of the mass distribution.

So if you stay where you are now, the gravitational field doesn't change, and the local gravitational acceleration doesn't change. If you were to move to the new position of the surface, the field would increase.

The value of acceleration due to gravity varies inverse squarely as the distance from the center of earth to the center of gravity of the object. Suppose we are concerned about the value of acceleration due to gravity on the earth's surface. Then the distance between the object and the center of earth will be the radius of the earth (assuming the earth to be spherical in shape) R. Let the corresponding value be g (which is approximately 9.8 m/s^2). Now keeping the mass constant and we try to reduce the volume (i.e, we are making the earth more dense as decrease in radius of earth corresponds to a decrease in volume of earth by a factor of r^3, where r is the change in radius, but. since we kept the mass constant, the density increases).
Now let the decreased new radius is R'. Since the new radius is less than the old (R' This tells us why neutrino stars, black holes etc have so much gravitational pull. These celestial objects are very dense. As in the case of a black hole, it contains a very huge massive core compressed highly to a very small volume (imagine the entire solar system compressed to to a volume of Texas city!!!!!!).
Remember what we have discussed here is the effect of acceleration due to gravity at the earth' surface on shrinking the volume of earth keeping the mass of earth constant. As you flew away to high altitudes you need to add up the altitude also to the radius of the earth in the denominator. On the other hand, if you are somewhat inside the earth, you cannot apply the mass of earth to the equation. Suppose you have went about 3 km down to earth. Then you need to apply the mass surrounding a spherical volume of radius (R-3) km in your equation with R'=(R-3) km.
where R and R' are in km. At the center of earth, the radius will be zero and one will possibly imagine infinite acceleration due to gravity at the center. But the mass is also zero there (since you are considering a single point). So you will get 0/0, which in this case on applying the limit gives you a finite value. But the force acting on that object will be zero. It's because due to spherical symmetry of the object with other parts of earth, the forces of attraction acting in all directions with the same magnitude cancels each other, resulting in a net force of zero. i.e, the object feels no weight at all.

The acceleration due to gravity on the surface of Earth can change in various ways as the earth shrinks.

1. If you look up at the rate of change of g with respect to decrease in R the radius of Earth, it will vary as 1/Radius of the old value ;

so the acceleration due to gravity increases with reduction in radius.

2. If you also consider the change in spin of the Earth as radius decreases the spinning velocity will increase due to shrinkage .

Therefore the centrifugal pull of the earth will increase and then the g(effective) value should decrease-

as the correction due to spin velocity is not substantial the earlier rate of increase(in 1) due to reduction in R will predominate over it and

a Net increase in g-value is expected on the surface of the earth