1. (i) r has been increased by a factor of
3 from the centre of planet. C1
g = (40/32 =) 4.4(4)
(N kg–1) A1
(ii) M
= gr2 / G
M = (40 × [2.0 × 107]2) / 6.67 × 10–11 C1
M = 2.4 × 1026 (kg) A1
(iii)
M
=
ρV = 4/3
πr3 ρ M1
g =
GM /
r2 
r3 /
r2 (Hence
g r) A1
[6]
2 The astronaut is accelerating / has
centripetal acceleration (1)
and the space station has the same acceleration (1)
a person does not feel gravity (1)
only feels forces applied by contact with the walls of the space station
(1)
no support force from the space station (as they have the same
acceleration) (1) 4
MAXIMUM
(4)
[4]
3. Period
24 hours (1)
Satellite must stay locked into Earth’s period of rotation (or wtte) (1)
Plane Equatorial (1)
Centre of orbit must be centre of Earth because axis of orbit must be spin axis
of Earth (1)
(gravitational force above equator is only force available to provide
centripetal force in a synchronised orbit, otherwise an engine is required)
Direction Same as Earth’s rotation (1)
(otherwise satellite and Earth would counter rotate)
[5]
4. (a) i. F = GMm/r2 or F α Mm/r2 with labels (1) 1
ii. finite universe contracts/ resultant force on stars (1) 1
(b) Any
2 from
i. (satellite B) has larger circumference/smaller velocity
(satellite B) Gravitational field strength is less
(satellite B) Centripetal force is less 2
ii.(accept calculation from either
satellite)
r13/
T12 = r23/ T22 (1)
satellite A satellite
B
r23 = 70003 × 57.22 / 1.632 r23 = 671003 × 57.22 / 1.632 (1)
r2 = 75,030 km r2 = 75, 320 km (1) 3
(= 75,000 km) (=
75,000 km)
(c) Land-based
are (any 3) 1 mark for each
more light can be collected/ made larger
more stable
more manoeuvrable
cheaper to build/repair
longer lifetime/ not exposed to high velocity particles
greater access 3
[10]
8. (i) 4.5 (N kg–1) 1
(ii) g = (–)GM/r2 1
(iii) g ∞ 1/r2;so value is 40/9 = 4.4(4)
(N kg–1) ecf c(i) 2
[4]
×
2.0 × 107/(4.5 × 103) = 2.8 × 104 m s–1 (2)
½ mv2; = 0.5 × 1500 × (2.8 × 104)2 = 5.9 × 1011 (J) (2) 4
aliter: F = mv2/R; = mg; so ½ mv2 = ½ mgR;= 6.0 × 1011 (J)
[8]
