ON 2008 / P4 / Q1



F = GMm/R2
FC = mv2/R = mω2R2/R = mω2R
GMm/R2 - mω2R
The value of R in the term “mω2R” varies from maximum to zero
/minimum as it moves from equator to the pole(s) respectively.
ON 2008 / P4 / Q1
A real planet is not a perfect sphere, so its radius R varies as g ∞1/R2
If the density of the planet varies throughout its volume.
MJ 2008 / P4 / Q1
It is the angle subtended at the centre of a circle by an arc equal to the
radius of the circle.
The angle swept by the small mass about point P in unit time is defined
as the angular speed.
MJ 2008 / P4 / Q1
Fc = Ff [ Fictional force provides the centripetal force]
Mω2d = 0.72W
ω2 = 0.72Mg / Md
ω = 2 πn / t
ω = √0.72 x 9.81 / 0.35
n = ωt/2π = 4.49 x 60 / 2 x 3.141
= 4.49 rads-1
= 42.9 = 43
43 min-1
Fc = Mω2r suggests that the need for the centripetal force to stay on the
plate is more for mud at the edge than near the centre. Therefore, upon
slowly increasing the angular speed, the mud near the edge will first fly off
the plate.
ON 2010 / P42 / Q1
Gravitational field strength at a point is the gravitational force exerted per unit mass
on a small mass object placed at that point.





ON 2010 / P42 / Q1
At that point, a point mass will feel equal and opposite forces from the
Earth and the Moon. So, the resultant force as well as the resultant field
strength will be zero at that point.
gE = gM (= same )
GME/x2 = GMM/(60RE-x)2
Or, x2 /(60RE-x)2 = ME / MM = 6.00 x 1024 / 7.40 x 1022 = = 81.081
Or, x / 60RE - x =√81.081 = 9
Or, 10x = 540RE
Or, x = 54RE
54

MJ 2016 / P43 / Q1
From the definition we know that at infinity the gravitational potential is zero.
So, the gravitational potential at a point near a planet is negative since work has to
be done by the attractive force in moving a point mass from infinity to that point.

Φ = EP/m = mgh/m = gh
E=mΦ
ΔE=mΔΦ
MJ 2016 / P43 / Q1

ΔE=mΔΦ
= 180 x (-1 - (-14)) x 108
= 7.2 x 108 J [ +’ve]
7.2 x 108
increase
½ mv2 = m Δ Φ
V2 = 2 Δ Φ
V=√2ΔΦ
= √ 2 x (10- 4.4) x 108
= 3.3 x 104 ms-1
3.3 x 104
MJ 2015 / P42 / Q4

Simple harmonic motion is a motion where the displacement is directly
proportional to the acceleration however they are in opposite directions.
1.7 cm
0.2s
1.2s equals 2π rad
0.2s “
(2π x 0.2)/1.2 = 1.05 rad
1.05
a0 = - ω2x0
= (-) (2 π/1.2)2 x 0.030 [ T = 1.2 s]
= (-)0.82 ms-2
(-) 0.82
MJ 2015 / P42 / Q4

Maximum kinetic energy E = ½ mv2 = ½ m ω2x02
E ∞ x02 So, EQ/EP = xQ2/xP2 = 3.02/1,72 = 3.1
3.1


ON 2016 / P42 / Q4
mass
2 cm
15 cm at Equilibrium
mass
2 cm
2 cm
mass






0.225
0.525
ON 2016 / P42 / Q4
ω = 2π/T
= 2 x 3.141 / 0.3 = 20.94 = 21 rads-1 C1
21
vmax = ωx0 C1
= 20.94 x 2 x 10-2
= 0.418
= 0.42 ms-1




A1