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.
The acceleration of gravity g on the Earth surface is g = G M/R^2.
Using the parameters given, this is g = (4/3) pi G R rho, with
rho being the Earth density, and comes out to 9.824961 m/s^2.
V_cone = pi r^2 h/3
With height h = 2 r, V_cone = (2/3) pi r^3.
Setting planet masses equal:
M_cone = M
rho V_cone = rho V_earth
rho (2/3) pi r^3 = rho (4/3) pi R^3
r^3 = 2 R^3
r = 2^(1/3) R
The center of mass, COM, of a cone is at a point along the
axis 1/4 the height from the base. Calling ga and gb the apex and
base accelerations on the cone, these occur at distances
ra = (3/4) h = (3/2) r and rb = (1/4) h = (1/2) r from the COM.
So, ra = (3/2) 2^(1/3) R and rb = (1/2) 2^(1/3) R
ga = R^2/ra^2 g = (2/3)^2 2^(-2/3) g = 2.7508 m/s^2
gb = R^2/rb^2 g = 4 2^(-2/3) g = 24.7574 m/s^2
I note that my method and results were the same as Kenny M's.
We did start with slightly different values of g; I used
the values from the ideal Earth in the problem and he used
one from the literature, but the differences are not apparent
when looking at 3 significant figures at the end.
Edited on December 17, 2024, 8:51 pm
soln
