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: ΔEP = 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
{ΔEP = mgΔh}
ΔEP = 2.4 × 3.7
× 1800
ΔEP = 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
No comments:
Post a Comment
If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation