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Rasek [7]
3 years ago
15

Consider a satellite in a circular orbit around the earth. Why is it important to give a satellite a horizontal speed when placi

ng it in orbit? What will happen if the horizontal speed is too small? What will happen if the speed is too large?
Physics
1 answer:
Andrei [34K]3 years ago
4 0

Answer:

In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero,

Explanation:

When a satellite is in orbit the most important force is the docking of gravity with the Earth

             F = m a

where the acceleration is centripetal and F is the force of universal attraction

centripetal acceleration is

             a = v² / r

             

             F = m v² / r

In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero, the force also drops to serious and the satellite steels to Earth.

The speed of the satellite is provides the speed, by local for smaller speeds in satellite, it descends in its orbits and when the speed is amate you have the energy to stop an orb to go to a higher orbit.

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Answer:

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As per the question:

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\hat{A} = \frac{4.4\hat{i} + 4.4\hat{j} + 4.4\hat{k}}{\sqrt{()4.4}^{2} + (4.4)^{2} + (4.4)^{2}}

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\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}

(b) To calculate the angle between the two vectors say A and A' is given by:

\vec{A}\cdot \vec{A'} = \vec{A}\vec{A'}cos\theta                      

\theta = cos^{- 1}(\frac{\vec{A}\cdot \vec{A'}}{\vec{A}\vec{A'}})        (1)

Now,

The coordinates of the diagonally opposite corner, A' is <0, 0, 1> cm

Thus

\vec{A'} = 0\hat{i} + 0\hat{j} + 1\hat{k} = \hat{k}

Now,

Using equation (1) :

\theta = cos^{- 1}(\frac{(\frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}})\cdot \hat{k}}{|A||A'|})

|A||A'| = (\sqrt{4.4^{2} +4.4^{2} + 4.4^{2}})(\sqrt{0^{2} + 0^{2} + 0^{2}}) = 7.261

Thus

\theta = cos^{- 1}(\frac{\frac{1}{\sqrt{3}}}{7.261})

\theta = cos^{- 1}(0.07946) = 85.44^{\circ}

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