A small, solid cylinder with mass = 20 kg and radius = 0.10 m starts from rest and rotates without friction about a fixed axis t
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(a)
At a distance r=R from the centre of the planet, there is no effect due to the outer shell: so, the gravitational field strength at r=R is only determined by the gravity produced by the core of the planet.
So, the strength of the gravitational field is given by
where
G is the gravitational constant
M = 6.24 × 10^24 kg is the mass of the core of the planet
R = 4.11 × 10^6 m is the radius of the core
Substituting into the equation, we find
(b)
at distance r=3R from the centre, the particle feels the effect of gravity due to both the core of the planet and the outer shell between R and 2R.
So, we have to consider the total mass that exerts the gravitational attraction at r=3R, which is the sum of the mass of the core (M) and the mass of the shell (4M):
M' = M + 4M = 5M
Therefore, the gravitational acceleration at r=3R will be
And susbstituting
g = 24.6 m/s^2
found in the previous part, we find
From the concept of optics on a curvature of a spherical mirror, the proportion for which the focal length is equivalent to half the radius of curvature is fulfilled. Mathematically this is
Here,
f = Focal Length
R = Radius
Rearranging to find the radius we have,
Replacing with our values,
R = 2(13.8cm)
R = 27.6cm
Therefore the radius of the spherical surface from which the mirror was made is 27.6cm