Question

Planet-X has a mass of 3.42×1024 kg and a radius of 8450 km. What is the First Cosmic Speed i.e. the speed of a satellite on a low lying circular orbit around this planet? (Planet-X doesn't have any atmosphere.) Submit Answer Tries 0/12 What is the Second Cosmic Speed i.e. the minimum speed required for a satellite in order to break free permanently from the planet? Submit Answer Tries 0/12 If the period of rotation of the planet is 17.1 hours, then what is the radius of the synchronous orbit of a satellite?

Answer 1

Answer:

First cosmic speed = 5195.74m/s

Second cosmic speed = 7346.05m/s

The raduis of the synchronous 0rbit of satellite is 2.80×10^7m

Explanation:

The first cosmic speed Is determined using the Orbital speed equation which is given by:

V = Sqrt(GM/r)

Where G = gravitational constant = 6.67 ×10^-11

M = Mass of planet

r = radius of the planet

V = Sqrt (6.67×10^-11)(3.42×10^24)/(8450×10^3)

V = Sqrt (2.28×10^14)/(8450×10^3)

V = Sqrt ( 26995739.64)

V = 5195.75m/s

The second cosmic speed is given by :

V = Sqrt(2 × GM)/r

V = Sqrt (2 × (6.67×10^-11)(3.42×10^24)/(8450×10^3)

V = Sqrt( 4.5×10^14)/ (8450×10^3)

V = Sqrt(53964497.04)

V = 7346.05m/s

The raduis of the synchronous orbit if the satellite around the planet is given by:

r = Cuberoot(T^2GM/4 pi r^2where T is the period of rotation of the planet in second

Given :

T = 17.1 hours converting to seconds

T = 17.1 ×60 ×60 = 61560 seconds

Substituting into the equation

r = Cuberoot ([(61560)^2×(6.67×10^-11)(3.42×10^24)/ (4 ×3.142×r^2)]

r = 2.80×10^7m

Answer 2

Answer:

First cosmic speed  = 5200 m/s

Second cosmic speed = 7350 m/s

Radius of the synchronous orbit of a satellite = 28,000 km or $2.80\times10^7\ \text{m}$

Explanation:

The first cosmic speed is given by

$V_1 = \sqrt{\dfrac{GM_p}{R_p}}$

G is the universal gravitational constant with value $6.674\times10^{-11}\ \text{Nm}^2\text{/kg}^2$

$M_p$ is the mass of the planet; and

$R_p$ is the radius of the planet.

$V_1 = \sqrt{\dfrac{(6.674\times10^{-11}\ \text{Nm}^2\text{/kg}^2)(3.42\times10^{24}\text{ kg})}{(8.45\times10^6\ \text{m})}} = 5200\ \text{m/s}$

The second cosmic speed is given by

$V_2 = \sqrt{\dfrac{2GM_p}{R_p}} = \sqrt{2}V_1$

$V_2 = \sqrt{2}\times 5200\text{ m/s} = 7350\ \text{m/s}$

The radius of the synchronous orbit of a satellite around the planet is given by

$r = \sqrt[3]{\dfrac{T^2GM_p}{4\pi^2}}$

where T is the period of rotation of the planet in seconds.

Substituting known values,

$r = \sqrt[3]{\dfrac{(17.1\times60\times60\ \text{s})^2(6.674\times10^{-11}\ \text{Nm}^2\text{/kg}^2)(3.42\times10^{24}\text{ kg})}{4\pi^2}}$

$r = 2.80\times10^7\ \text{m}$