Question

Planet-X has a mass of 4.74×1024 kg and a radius of 5870 km. The First Cosmic Speed i.e. the speed of a satellite on a low lying circular orbit around this planet is 7.34 km/s.
1. What is the Second Cosmic Speed i.e. the minimum speed required for a satellite in order to break free permanently from the planet?
2. If the period of rotation of the planet is 16.6 hours, then what is the radius of the synchronous orbit of a satellite?

Answer 1

10,378.82 m/s is the second cosmic speed.

69,801 km is the radius of the synchronous orbit of a satellite.

Given

Mass of planet = 4.74 × $10^{24}$ kg

Radius of planet = 5870 km = 5870000m

First Cosmic speed = 7.34 km/sec

1) Second cosmic speed i.e. the minimum speed required for a satellite to break free permanently from the planet is also known as the escape velocity of a satellite.

It can be calculated by

v = √2GM/r where,

v= Escape velocity of the satellite

G = Gravitational constant

M = Mass of planet

r = Radius of planet

v = √[( 2 x 6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]

v = 10,378.82 m/s

2) Speed of the satellite at the given period

v = 2πr/T where,

T= Time period of rotation = 16.6 × 3600 seconds

r = v×T/2π

r = (7,338.93 x 16.6 x 3600 s) / (2π)

r = 69,801 km

Hence

The Second Cosmic Speed i.e. the minimum speed required for a satellite to break free permanently from the planet is 10,378.82 m/s.

And the radius of the synchronous orbit of a satellite is 69,801 km.

Learn more about cosmic speed here brainly.com/question/15351190

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