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kykrilka [37]
3 years ago
9

A satellite was in two separate crashes. In both crashes, the satellite had the same mass. Engineers want to know about the spee

d and direction of the satellite after the crashes. Why would the crash affect the motion of the satellite, and which crash caused a greater change in motion for the satellite?
WILL GIVE BRAINLIEST

Physics
1 answer:
Komok [63]3 years ago
5 0

Answer:

The fastest satellite must change orbit

The most massive body (m₁) transfers more momentum to the satellite,

Explanation:

For this problem we consider a system formed by the satellite and each of the bodies with which it collides, in this system the forces during the collision are internal, the amount of movement must be conserved. Let's write the momentum is two instants

Most massive body (m1)

initial. Before the crash

      p₀₁ = M v + m₁ v₁

after the crash

      p_{f1} = M v´ + m₁ v₁´

how momentum is conserved

     p₀ = p_{f}

Lighter body (m2)

      p₀₂ = M v + m₂ v₂

       p_{f2} = M v´ + m₂ v₂´

           

Let's clarify that the speed of the satellite and the object do not have the same direction, in general these shocks are elastic.

We can see that  p₀₁> p₀₂

Let us analyze the two cases when the body collides, The most massive body (m₁) transfers more momentum to the satellite, therefore there must be a greater change in its momentum and velocity.

The fastest satellite must change orbit, thus rotating at a different distance from Earth

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Hence 1st option is correct!
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Read 2 more answers
Planet-X has a mass of 4.74×1024 kg and a radius of 5870 km.
Nookie1986 [14]

(a) The speed of a satellite on a low lying circular orbit around this planet is  7,338.93 m/s.

(b) The minimum speed required for a satellite in order to break free permanently from the planet is 10,378.82 m/s.

(c) The radius of the synchronous orbit of a satellite is 69,801 km .

<h3>Speed of the satellite</h3>

v = √GM/r

where;

  • M is mass of the planet
  • r is radius of the planet

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

v = 7,338.93 m/s

<h3>Escape velocity of the satellite</h3>

v = √2GM/r

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

v = 10,378.82 m/s

<h3>Speed of the satellite at the given period </h3>

v = 2πr/T

r = vT/2π

r = (7,338.93 x 16.6 x 3600 s) / (2π)

r = 69,801 km

Thus, the speed of a satellite on a low lying circular orbit around this planet is  7,338.93 m/s.

The minimum speed required for a satellite in order to break free permanently from the planet is 10,378.82 m/s.

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

Learn more about minimum speed here: brainly.com/question/6504879

#SPJ1

4 0
1 year ago
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