The planet Jupiter and one of its moons, Io, may be considered to be uniform spheres that are isolated in space.

 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]

An isolated spherical planet has a diameter of 6.8 × 106 m. Its mass of 6.4 × 1023 kg may be assumed to be a point mass at the centre of the planet.

Question 10

An isolated spherical planet has a diameter of 6.8 × 10⁶ m. Its mass of 6.4 × 10²³ kg may be assumed to be a point mass at the centre of the planet.

(a) Show that the gravitational field strength at the surface of the planet is 3.7 N kg^-1. [2]

(b) A stone of mass 2.4 kg is raised from the surface of the planet through a vertical height of 1800 m.

Use the value of field strength given in (a) to determine the change in gravitational potential energy of the stone.

Explain your working. [3]

(c) A rock, initially at rest at infinity, moves towards the planet. At point P, its height above the surface of the planet is 3.5 D, where D is the diameter of the planet, as shown in Fig. 1.1.

Fig. 1.1

Calculate the speed of the rock at point P, assuming that the change in gravitational potential energy is all transferred to kinetic energy. [4]

Reference: Past Exam Paper – November 2014 Paper 41 & 42 Q1

Solution:

(a)

Gravitational field strength, g = GM / R²

Gravitational field strength, g = (6.67×10^–11) × (6.4×10²³) / (3.4×10⁶)²

Gravitational field strength, g = 3.7 N kg^-1

(b)

Change in gravitational potential energy: ΔE_P = mgΔh

because (the vertical height is much smaller than the radius) Δh ≪ R (or 1800 m ≪ 3.4 × 10⁶ m), the gravitational field strength g is constant

{ΔE_P = mgΔh}

ΔE_P = 2.4 × 3.7 × 1800

ΔE_P = 1.6 × 10⁴ J

(c)

{Gravitational potential ϕ = - GM / x

Gravitational potential energy = mϕ = - GMm / x}

Gravitational potential energy = (–) GMm / x

{x is the separation between the rock and the CENTRE of the planet (since the planet is considered to be a point mass, all its mass is concentrated at the centre).}

{The GPE is all transferred to kinetic energy. So, considering the magnitudes,

½ mv² = GMm / x

giving v² = 2GM / x.}

v² = 2GM / x

{The rock is at a distance 3.5D from the surface of the planet. The surface is itself at a distance of D/2 = 0.5D from the centre of the planet.

So, separation x = 3.5D + 0.5D = 4D}

Separation x = 4D = 4 × 6.8×10⁶ m

{v² = 2GM / 4D}

v² = 2 × (6.67×10^–11) × (6.4×10²³) / (4 × 6.8×10⁶)

v² = 3.14 × 10⁶

Speed v = 1.8 × 10³ m s^–1