12 ORBITAL MOTION 12.6 Planetary orbits

!

× × ×

2

= 1.107 10 m.

Worked example 12.2: Acceleration of a rocket

Question: A rocket is located a distance 3.5 times the radius of the Earth above

the Earth’s surface. What is the rocket’s free-fall acceleration?

Answer: Let R⊕ be the Earth’s radius. The distance of the rocket from the centre

of the Earth is r 1 = (3.5 + 1) R⊕ = 4.5 R⊕. We know that the free-fall acceleration

of the rocket when its distance from the Earth’s centre is r 0 = R⊕ (i.e., when it is

at the Earth’s surface) is g 0 = 9.81 m/s^2. Moreover, we know that gravity is an

inverse-square^ law^ (i.e.,^ g^ ∝^ 1/r^2 ).^ Hence,^ the^ rocket’s^ acceleration^ is^

g 1 = g 0

r (^0 2) =
r 1
9.81 × 1
= 0.484 m/s^2.
(4.5)^2
Worked example 12.3: Circular Earth orbit
Question: A satellite moves in a circular orbit around the Earth with speed v =
6000 m/s. Determine the satellite’s altitude above the Earth’s surface. Determine
the period of the satellite’s orbit. The Earth’s mass and radius are M⊕ = 5.97 ×
1024 kg and R⊕ = 6.378 × 106 m, respectively.
Answer: The acceleration of the satellite towards the centre of the Earth is v^2 /r,
where r is its orbital radius. This acceleration must be provided by the accelera-
tion G M⊕/r^2 due to the Earth’s gravitational attraction. Hence,
v

G M⊕
.
r r^2
The above expression can be rearranged to give
r =
G M⊕

v^2
(6.673 10 −^11 ) (5.97 1024 ) 7
(6000)^2
×
Thus, the satellite’s altitude above the Earth’s surface is
h = r − R⊕ = 1.107 × 107 − 6.378 × 106 = 4.69 × 106 m.