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Diano4ka-milaya [45]
2 years ago
9

Sputnik, the first artificial satellite to orbit the Earth, had a mass of 83.6kg and travelled at 7574 m/s. The radius of the ea

rth is 6371 km and its mass is 5.972 x 10^24 kg. What gravitational force did Sputnik apply to the earth during this orbit?
Physics
1 answer:
konstantin123 [22]2 years ago
5 0

Answer:

690.6 N

Explanation:

The gravitational force that the Earth exerts on Sputnik acts as centripetal force to keep the satellite in motion, so we can write

\frac{GMm}{r^2}=m\frac{v^2}{r}

where

G is the gravitational constant

M = 5.972 x 10^24 kg is the Earth's mass

m = 83.6 kg is the satellite's mass

v = 7574 m/s is the satellite's speed

r is the distance of the satellite from the Earth's center

Solving for r, we find

r=\frac{GM}{v^2}=\frac{(6.67\cdot 10^{-11})(5.972\cdot 10^{24})}{(7574)^2}=6.944\cdot 10^6 m

Now we can find the gravitational force exerted by the Earth on Sputnik:

F=\frac{GMm}{r^2}=\frac{(6.67\cdot 10^{-11})(5.972\cdot 10^{24})(83.6)}{(6.944\cdot 10^6)^2}=690.6 N

And according to Newton 3rd law (action-reaction, the force exerted by Sputnik on the Earth is equal to the force exerted by the Earth on Sputnik, so this is the value of the force we are requested to find.

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Andrea and Chuck are riding on a merry-go-round. Andrea rides on a horse at the outer rim of the circular platform, twice as far
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a) Their angular speeds are the same

b) Andrea's tangential speed is twice the value of Chuck's tangential speed

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a)

The angular speed of Andrea and Chuck is the same.

Let's call \omega the angular speed at which the merry-go-round is rotating. We know that the angular speed is defined as:

\omega= \frac{2\pi}{T}

where

2 \pi is the angular displacement covered in one revolution

T is the period of revolution

The merry go round is a rigid body, so all its point cover the same angular displacement in the same time: this means that it doesn't matter where Andrea and Chuck are located along the merry-go-round, their angular speed will still be the same.

b)

For an object in circular motion, the tangential speed is given by

v=\omega r

where

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r is the distance from the centre of rotation

Here let's call r_c the distance at which Chuck is rotating, so his tangential speed is

v_c = \omega r_c

Now we know that Andrea is rotating twice as far from the centre, so at a distance of

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So his tangential speed is

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So, Andrea's tangential speed is twice the value of Chuck's tangential speed.

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

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