Based on the calculations, the speed required for this satellite to stay in orbit is equal to 1.8 × 10³ m/s.
<u>Given the following data:</u>
- Gravitational constant = 6.67 × 10⁻¹¹ m/kg²
- Mass of Moon = 7.36 × 10²² kg
- Distance, r = 4.2 × 10⁶ m.
<h3>How to determine the speed of this satellite?</h3>
In order to determine the speed of this satellite to stay in orbit, the centripetal force acting on it must be sufficient to change its direction.
This ultimately implies that, the centripetal force must be equal to the gravitational force as shown below:
Fc = Fg
mv²/r = GmM/r²
<u>Where:</u>
- m is the mass of the satellite.
Making v the subject of formula, we have;
v = √(GM/r)
Substituting the given parameters into the formula, we have;
v = √(6.67 × 10⁻¹¹ × 7.36 × 10²²/4.2 × 10⁶)
v = √(1,168,838.095)
v = 1,081.13 m/s.
Speed, v = 1.8 × 10³ m/s.
Read more on speed here: brainly.com/question/20162935
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Answer:
you have probably missed some details in the question.
The answer is a
Hope this helps!
Answer:
Angle:
Explanation:
<u>Two-Dimension Motion</u>
When the object is moving in one plane, the velocity, acceleration, and displacement are vectors. Apart from the magnitudes, we also need to find the direction, often expressed as an angle respect to some reference.
Our boy can swim at 3 m/s from west to east in still water and the river he's attempting to cross interacts with him at 2 m/s southwards. The boy will move east and south and will reach the other shore at a certain distance to the south from where he started. It happens because there is a vertical component of his velocity that is not compensated.
To compensate for the vertical component of the boy's speed, he only has to swim at a certain angle east of the north (respect to the shoreline). The goal is to make the boy's y component of his velocity equal to the velocity of the river. The vertical component of the boy's velocity is
where is the speed of the boy in still water and is the angle respect to the shoreline. If the river flows at speed , we now set
An electron is emitted in both positive and negative beta decay, although the positive one is called positron emission.