A Classical Approach of Newtonian Mechanics

12 ORBITAL MOTION 12.6 Planetary orbits

× × ×

× × × ×

= −7.495 10 J.

Answer: Let ω be the planet’s orbital angular velocity. The planet accelerates

towards the star with acceleration ω^2 r. The acceleration due to the star’s gravi-

tational attraction is G M∗/r^2 , where M∗ is the mass of the star. Equating these
accelerations, we obtain

Now,

ω^2 r =

G M∗
.
r^2
2 π
T =.
ω
Hence, combining the previous two expressions, we get

4 π^2 r^3

M∗ =

G T 2

.

Thus, the mass of the star is

M∗ =

4 π^2 (2.34 1011 )^3 30

(6.673 × 10 −^11 ) × (4.3 × 107 )^2

= 4.01 × 10

kg.

Worked example 12.6: Launch energy

Question: What is the minimum energy required to launch a probe of mass m =
120 kg into outer space? The Earth’s mass and radius are M⊕ = 5.97 × 1024 kg

and R⊕ = 6.378 × 106 m, respectively.

Answer: The energy which must be given to the probe should just match the

probe’s gain in potential energy as it travels from the Earth’s surface to outer
space. By definition, the probe’s potential energy in outer space is zero. The

potential energy of the probe at the Earth’s surface is

U = −

G M⊕ m

=

R⊕

(6.673 10 −^11 ) (5.97 1024 ) 120 9

(6.378 × 106 )

×

Thus, the gain in potential energy, which is the same as the minimum launch

energy, is 7.495 × 109 J.