Question 11
(a)
State Newton’s law of gravitation. [2]
(b)
The planet Jupiter and one of its moons, Io,
may be considered to be uniform spheres that are isolated in space.
Jupiter has radius R
and mean density ρ.
Io
has mass m and is in a circular
orbit about Jupiter with radius nR, as illustrated in
Fig. 1.1.
Fig. 1.1
The time for Io
to complete one orbit of Jupiter is T.
Show that the time T
is related to the mean density ρ
of Jupiter by the expression
ρT2 = 3πn3
/ G
where G
is the gravitational constant. [4]
(c)
(i) The radius R of Jupiter is 7.15 ×
104 km and the distance between the centres of
Jupiter and Io is 4.32 × 105 km.
The period T
of the orbit of Io is 42.5 hours.
Calculate the mean
density ρ of Jupiter. [3]
(ii)
The Earth has a mean density of 5.5 × 103 kg m-3. It is said to be a planet
made of rock.
By reference to your
answer in (i), comment on the possible
composition of Jupiter. [1]
[Total: 10]
Reference: Past Exam Paper – November 2017 Paper 42 Q1
Solution:
(a)
Newton’s law of gravitation states that the
gravitational force between masses is proportional to the product of the
point masses and inversely proportional to the square of their
separation.
(b)
{Density = mass / volume
Mass = Density × Volume
Since Jupiter is considered to be a uniform
sphere, its volume = 4/3 πr3
}
mass of Jupiter (M) = (4 / 3) πR3 ρ
{For
the centripetal force, we may either use angular velocity or velocity.
Speed =
distance / time = circumference / period}
ω = 2π / T or v = 2πnR / T
{The gravitational force provides the
centripetal force.
Centripetal force = gravitational force}
(m)ω2x = GM(m) / x2 or (m)v2 / x = GM(m) / x2
{Note that x
= radius of orbit = nR
the mass of Jupiter is given in terms of R, its radius}
substitution and correct algebra leading to ρT2 =
3πn3 / G
(m)ω2x = GM(m) / x2
(2π / T)2 nR = G ×
((4 / 3) πR3 ρ) / n2R2
(2π / T)2 n3R3 = G × (4 / 3) πR3 ρ
ρT2 = 3πn3 / G
(c)
(i)
{ ρT2 = 3πn3 / G We need to find n
first.}
{R
= 7.15 × 104 km and nR = 4.32 × 105 km so, n =
nR / R}
n = (4.32 × 105)
/ (7.15 × 104) or n = 6.04
ρ × (42.5
× 3600)2 = (3π × 6.043) / (6.67 × 10-11)
ρ = 1.33 × 103 kg m-3
(ii)
{The density is approximately one fifth
of the density of the Earth.}
Jupiter likely to be a gas/liquid (at high pressure) [allow other sensible suggestions]
