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

A wire of cross-sectional area 1.5 mm2 and length 2.5 m has a resistance of 0.030 Ω. Calculate the resistivity of the material of the wire in nΩ m.

Question 27

(a) State two SI base units other than kilogram, metre and second. [1]

(b) Determine the SI base units of resistivity. [3]

(c) (i) A wire of cross-sectional area 1.5 mm² and length 2.5 m has a resistance of 0.030 Ω.

Calculate the resistivity of the material of the wire in nΩ m. [3]

(ii)

1. State what is meant by precision.

2. Explain why the precision in the value of the resistivity is improved by using a

micrometer screw gauge rather than a metre rule to measure the diameter of the

wire.

 [2]

[Total: 9]

Reference: Past Exam Paper – June 2017 Paper 22 Q1

Solution:

(a) kelvin, mole, ampere, candela

any two

(b)

{Resistance of wire: R = ρL / A

Resistivity ρ = RA / L

Now, to determine the unit of resistance R.

Power: P = I²R

R = P / I²

Power = energy / time = force × distance / time = mas / t

Units of power: kg ms^-2 m s^-1 = kg m² s^-3 

Unit of current = A

Units of R: kg m² s^-3 A^-2

Resistivity ρ = RA / L

Unit of resistivity: kg m² s^-3 A^-2 m² m^-1 = kg m³ s^-3 A^-2

}

(c) (i)

ρ = (RA / L)                                        

{The area should be in m². To convert from mm² to m², multiply by 10^-6 (that is, divide by 1 000 000).}

ρ = (0.03 × 1.5×10^-6) / 2.5 (= 1.8 × 10^-8)

ρ = 18 nΩ m                                       

(ii)

1. Precision is determined by the range in the measurements

2. A metre rule measures to ± 1 mm while a micrometer measures up to ± 0.01 mm. So, there is less percentage uncertainty.