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]

Two small solid metal spheres A and B have equal radii and are in a vacuum. Their centres are 15 cm apart.

 Question 30

Two small solid metal spheres A and B have equal radii and are in a vacuum. Their centres are 15 cm apart.

Sphere A has charge +3.0 pC and sphere B has charge +12 pC. The arrangement is illustrated in Fig. 5.1.

 

Fig. 5.1

 

Point P lies on the line joining the centres of the spheres and is a distance of 5.0 cm from the centre of sphere A.

 

(a) Suggest why the electric field strength in both spheres is zero. [2]

 

 

(b) Show that the electric field strength is zero at point P. Explain your working. [3]

 

 

(c) Calculate the electric potential at point P. [2]

 

 

(d) A silver-107 nucleus (¹⁰⁷47 Ag) has speed v when it is a long distance from point P.

 

Use your answer in (c) to calculate the minimum value of speed v such that the nucleus can reach point P. [3]

[Total: 10]

 

 

 

Reference: Past Exam Paper – November 2016 Paper 41 & 43 Q5

 

 

 

Solution:

(a)

In an electric field, the charges (in a conductor) would move.

Since there is no movement of charge inside the sphere, the field strength is zero.

 

 

(b)

For the electric field strength to be zero at P, the fields due to the charged spheres must be (equal and) opposite in direction, so that E = 0.

 

{Showing that the field strength due to each sphere are equal in magnitude:

Electric field strength, E = Q / 4πϵ₀r²

            where r is the distance of P from the centre of the sphere

            For sphere A, r = 5 cm = 0.05 m

            For sphere B, r = 15 – 5 = 10 cm = 0.10 m

 

Note: 1 pC = 1×10^-12 C}

 

at P, EA = (3.0×10^-12) / [4 πϵ₀ × (5.0×10^-2)²] (= 10.79 N C^-1)            

at P, EB = (12×10^-12) / [4 πϵ₀ × (10×10^-2)²] (= 10.79 N C^-1)  

 

 

(c)

{Electric potential V = Q / 4πϵ₀r

At point P, the electric potential is due to both spheres A and B. Since electric potential is a scalar, we add both potentials directly (no directions are involved).

Electric potential at P = VA + VB

Electric potential at P = (Q_A / 4πϵ₀r_A) + (Q_B / 4πϵ₀r_B)  

Electric potential at P = 1/4πϵ₀ × [(Q_A / r_A) + (Q_B / r_B)]

 

1/4πϵ₀ = 8.99×10⁹ mF^-1          as given in the list of data in the question paper

 

Electric potential at P = 8.99×10⁹ × [(Q_A / r_A) + (Q_B / r_B)]}

 

potential = 8.99×10⁹ × {(3.0×10^-12) / (5.0×10^-2) + (12×10^-12) / (10×10^-2)}

potential = 1.62 V

                                                      

 

(d)

{Kinetic energy of nucleus = Electrical energy

Electrical energy = qV                        where q is the charge of the nucleus}

 

½mv² = qV

 

{The silver nucleus contains 107 nucleons. So, its mass is 107u (= 107 × 1.66×10^-27 kg).}

EK = ½ ×107 × 1.66×10^-27 × v²

 

{The silver nucleus contains 47 protons. So, its charge is 47e (=47 × 1.60×10^-19 C).}

qV = 47 × 1.60×10^-19 × 1.62   

v² = 1.37 × 10⁸

v = 1.2 × 10⁴ m s^-1                        

 

A rectangular block of steel supporting a very large component of a bridge has a height of 15 cm and a cross-section of 20 cm × 12 cm.

 Question 18

A rectangular block of steel supporting a very large component of a bridge has a height of 15 cm and a cross-section of 20 cm × 12 cm. It is designed to compress 1 mm when under maximum, evenly distributed, load.

 

The Young modulus of steel is 2.0 × 10¹¹ N m^-2.

 

What is the maximum load it can support?

A 32 MN                     B 56 GN                      C 720 GN                    D 32 TN

 

 

 

Reference: Past Exam Paper – November 2017 Paper 12 Q21

 

 

 

Solution:

Answer: A.

Length L = 15 cm = 0.15 m

Cross-sectional area A = (0.20 × 0.12) m²

Compression, e = 1 mm = 0.001 m

Young modulus E = 2.0×10¹¹ N m^-2

Maximum load F = ???

 

Young modulus E = stress / strain = FL / Ae

E = FL / Ae

F = EAe / L

F = (2.0×10¹¹ × 0.20×0.12 × 0.001) / 0.15

F = 32 000 000 N = 32×10⁶ N = 32 MN