Escape speed from the surface of earth is about

2 answers:

8 0

Escape velocity is the speed that an object needs to be traveling to break free of a planet or moon's gravity well and leave it without further propulsion. For example, a spacecraft leaving the surface of Earth needs to be going 7 miles per second, or nearly 25,000 miles per hour to leave without falling back to the surface or falling into orbit

dezoksy [38]3 years ago

6 0

On the surface of the Earth, the escape velocity is about 11.2 km/s, which is approximately 33 times the speed of sound (Mach 33) and several times the muzzle velocity of a rifle bullet (up to 1.7 km/s). However, at 9,000 km altitude in "space", it is slightly less than 7.1 km/s.

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koban [17]

Answer:

r = 4.24x10⁴ km.

Explanation:

To find the radius of such an orbit we need to use Kepler's third law:

$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}$

where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km.

From equation (1), r₁ is:

$r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}$

$r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}$

$r_{1} = 4.24 \cdot 10^{4} km$

Therefore, the radius of such an orbit is 4.24x10⁴ km.

I hope it helps you!

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