Answer:
1.64x[tex]10^{-4}[/tex]rad/s
Explanation:
Given:
satellite is in elliptical orbit with a period T= 7.20 x [tex]10^{4}[/tex] s
Mass of planet m= 7.00x [tex]10^{24}[/tex] kg
Satellite's angular speed ωa- = 4.987 10-5 rad/s
At aphelion, at radius Ra= 5.1 x 10^7 m
Assuming torque is zero for this system, therefore, the conservation of angular momentum can be defines as
Lp = La
Ip ωp = Ia ωa--->eq(1)
Where,
'a' represents aphelion and 'p' represents perihelion
ω represents angular velocity
and I represents the rotational inertia
Since, I= 1/2 mR²
Here R is the radius at aphelion / perihelion
m = mass of planet
eq(1)=> ωp= Ia ωa/ Ip
ωp= (1/2 m Ra² ωa) / 1/2 m Rp²
ωp= (Ra/Rp)² ωa --->eq(2)
In order to find Rp, we use : 2a= Rp + Ra
where a represents semimajor axis
with the help of Kepler's third law for elliptical orbits
a= ∛(GmT²/4π²)
a= ∛ 6.67x [tex]10^{-11}[/tex] x 7.00x [tex]10^{24}[/tex] x (7.20 x [tex]10^{4}[/tex])² / 4π²
a= 3.97 x [tex]10^{7}[/tex]m
2a= Rp + Ra
Rp= 2a-Ra
Rp= 2 x 3.97 x [tex]10^{7}[/tex]- 5.1x [tex]10^{7}[/tex]
Rp= 2.84 x [tex]10^{7}[/tex]m
Substituting all the required values in eq 2, we have
ωp= (Ra/Rp)² ωa
ωp= (5.1x [tex]10^{7}[/tex]/2.84 x [tex]10^{7}[/tex])² x 4.987 x [tex]10^{-5[/tex]
ωp= 3.224 x 4.987 x [tex]10^{-5[/tex]
ωp= 1.64x[tex]10^{-4}[/tex]rad/s
Therefore, angular speed at perihelion is 1.64x[tex]10^{-4}[/tex]rad/s
