Question

A satellite is in elliptical orbit with a period of 7.20 104 s about a planet of mass 7.00 1024 kg. At aphelion, at radius 5.1 107 m, the satellite's angular speed is 4.987 10-5 rad/s. What is its angular speed at perihelion?

Answer 1

Answer:

1.64x$10^{-4}$rad/s

Explanation:

Given:

satellite is in elliptical orbit with a period T= 7.20 x $10^{4}$ s

Mass of planet m=  7.00x $10^{24}$ 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 $10^{-11}$ x 7.00x $10^{24}$ x (7.20 x $10^{4}$)² / 4π²

a=  3.97 x $10^{7}$m

2a= Rp + Ra

Rp= 2a-Ra

Rp= 2 x 3.97 x $10^{7}$-  5.1x $10^{7}$

Rp= 2.84 x $10^{7}$m

Substituting all the required values in eq 2, we have

ωp= (Ra/Rp)² ωa

ωp= (5.1x $10^{7}$/2.84 x $10^{7}$)² x  4.987 x $10^{-5$

ωp= 3.224 x 4.987 x $10^{-5$

ωp= 1.64x$10^{-4}$rad/s

Therefore, angular speed at perihelion is 1.64x$10^{-4}$rad/s