Page 91 - text book form physics kssm 2020

Chapter 3
Gravitation
Figure 3.12 shows a satellite at height, h from the surface of the Earth. R is the radius of
the Earth and r is the distance of the satellite from the centre of the Earth, which is the radius
of the orbit.

Satellite
Sa
On the surface of the Earth,
At height, h from the surface of the h height, h = 0.
Earth, distance from the centre of r Th erefore,
the Earth is r = (R + h). r = radius of the Earth, R.
R R
With this, gravitational acceleration, Gravitational acceleration
GM on the surface of the Earth,
g = 2
(R + h) GM
g = 2 , where
Earth R
M is the mass of the Earth.

Figure 3.12 A satellite at height, h from the surface of the Earth
Example 1
24
6
Mass of the Earth is 5.97 × 10 kg and radius of the Earth is 6.37 × 10 m. Calculate
–2
2
gravitational acceleration on the surface of the Earth. [G = 6.67 × 10 –11 N m kg ]
Solution:
24
M = 5.97 × 10 kg
Step –11 2 –2
List the given information with symbols. 123 G = 6.67 × 10 N m kg
6
r = 6.37 × 10 m
Step g = GM
Identify and write down the formula used. 123 r 2

24
Step g = (6.67 × 10 –11 ) × (5.97 × 10 )
6 2
Make numerical substitution into the formula 14243 (6.37 × 10 )
and perform the calculations. = 9.81 m s –2

Example 2

A radar imaging satellite orbits around the Earth at a height of 480 km. What is the value of
gravitational acceleration at the position of the satellite?
–2
24
2
6
[G = 6.67 × 10 –11 N m kg , M = 5.97 × 10 kg, R = 6.37 × 10 m]
Solution:
Height of orbit, h = 480 km g = GM
= 480 000 m (R + h) 2
24
(6.67 × 10 –11 ) × (5.97 × 10 )
= 6 2
(6.37 × 10 + 480 000)
= 8.49 m s –2

85
85
85
3.1.3
3.1.3 85