Page 116 - text book form physics kssm 2020

Solution:

(i) Gravitational acceleration is calculated using the formula g = GM
R 2
Th e Moon
–11
(6.67 × 10 ) × (7.35 × 10 )
22
g =
6 2
(1.74 × 10 )
= 1.62 m s –2
Th e Sun
30
–11
(6.67 × 10 ) × (1.99 × 10 )
g =
8 2
(6.96 × 10 )
= 274.0 m s –2

2GM
(ii) Escape velocity is calculated using the formula v =
R
Th e Moon

–11
22
v = 2 × (6.67 × 10 ) × (7.35 × 10 )
6
1.74 × 10
3
= 2.37 × 10 m s –1
Th e Sun

2 × (6.67 × 10 ) × (1.99 × 10 )
30
–11
v =
(6.96 × 10 )
8
5
= 6.18 × 10 m s –1
• Th e Moon has low gravitational acceleration and escape velocity because the mass of the
Moon is smaller than that of the Sun.
• Th e Sun is the largest body in the Solar System. Gravitational acceleration on the Sun and
escape velocity from the Sun have the highest values compared with those of the Moon as
well as other planets.

Formative Practice 3.3
3.3

1. Compare and contrast geostationary and non-geostationary satellites.

2. What factors determine the linear speed of satellites orbiting the Earth?
3. State two factors which infl uence the value of escape velocity from a planet.
4. Discuss whether escape velocity from the Earth for spacecraft X of mass 1 500 kg is
diff erent from spacecraft Y of mass 2 000 kg.
5. Proba-1 satellite orbits the Earth at a height of 700 km. What is the linear speed of this
satellite?
24
2
–11
–2
[G = 6.67 × 10 N m kg , mass of the Earth = 5.97 × 10 kg, radius of the Earth =
6
6.37 × 10 m]
110
110 3.3.4
3.3.4