Suppose that, in a distant galaxy, there is a solar system in which, instead of being spheres, the planets are right circular cones (with heights equal to the diameters of their bases). Suppose one of these planets has the same total volume and mass as our Earth, but a uniform density.

What would be the gravitational acceleration on a person standing in the center of the circular base, and what would be the gravitational acceleration on a person standing at the apex?

Assume the Earth is a perfect sphere with a radius of 6,370 km and an average density of 5,518 kg/m^3. Use a value of 6.673×10-11 N m^2/ kg^2 for G. Express your answers correct to three significant figures.

 First collect and name some variables and get consistent units (kg, m, s):

G = 6.673x10^-11
Radus of the earth = RE = 6.370*10^6 m
Radius of cone planet = RC = TBD m
Density = p = 5.518 kg/m^3
Acceleration due to gravity on the surface of the earth (a scientific given) = 9.807 m/(s^2)

Isaac Newton determined that the force of gravity F = G*m1*m2/(r^2) where
G=above, m1=mass of the planet, m2=mass of the man, r=the distance between the center of masses of the two objects (man and planet).  We can ignore the height of the man when dealing with only three significant figures.

But we don’t need to calculate all this for both the earth and the cone planet because their masses and volumes and therefore avg. density are given as equal. Since F=ma, we really only need to figure out the distance of the man to the center of mass of the cone planet for the two cases and compare that properly to the radius of the earth which is given.  Specifically: (accel cone planet)/(accel earth) = (radius of the earth)^2/(man distance to center of mass of cone planet)^2  or  (accel cone planet)= (accel earth)*(RE^2)/(man distance to center of mass of cone planet)^2.

Specifically:

Volume of the earth (sphere) = Volume of the Cone Planet
4/3*(pi)*RE^3 = (pi)*(RC^2)*(2*RC)/3
Simplifying, RC = (2^(1/3))*RE.

One last thing. The center of mass of a right circular cone is on the axis of rotation at ¼ the total height from the base, or ¼ * (2*RC), or 0.5*RC.

CASE 1 �" Apex.  The distance of the man to the center of mass is ¾ (height of cone) = ¾*2*RC, or 1.5*RC.
Accel of cone planet in this case is 9.807/(1.5^2)/[2^(2/3)] = 2.7458 m/(s^2) or to three significant figures =2.75 m/(s^2).

CASE 2 �" on center of the base.  The distance of the man to the center of mass is 0.5*RC.  This simplifies to accel of the cone planet = 9.807*(2^(4/3)) = 24.712 m/(s^2) or to three significant figures 24.7 m/(s^2).

Edited on December 17, 2024, 4:21 pm

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Posted by Kenny M on 2024-12-17 15:48:46