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vlada-n [284]
3 years ago
8

The centripetal force acting on a satellite in orbit

Physics
2 answers:
krok68 [10]3 years ago
8 0

The centripetal force (of gravity) on a satellite in orbit is an
unbalanced force (because there's no equal force pulling
the satellite away from Earth), changes the direction of the
satellite (into a closed orbit instead of a straight line), and
always acts toward the center of whatever curve the satellite
happens to be on at the moment.

AysviL [449]3 years ago
3 0

Answer : (D) " all of the above "

Explanation :

The forces acting on a satellite are Earth's gravitational force and centripetal force. The properties of centripetal force acting on a satellite in orbit are as follows :

(1) It acts as an unbalanced force on the satellite. This is due to the reason that there is no same force present which pulls the satellite away from the earth.

(2) It changes the direction of the satellite. The satellite is revolving with a constant speed while the direction is kept on changing (in a circular motion).

(3) Centripetal force always acts towards the center.

So, the correct option is (D) " all of the above "

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The Correct answer would be D. Work equals force times distance.
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What are greenhouse gasses? How does carbon dioxide enter the atmosphere? (Site 1)
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To teach you how to find the parameters characterizing an object in a circular orbit around a much heavier body like the earth.
Karolina [17]

Answer:

a)  T² = (\frac{4\pi ^2}{GM})  r³

b) veloicity the dependency is the inverse of the root of the distance

kinetic energy  depends on the inverse of the distance

potential energy dependency is the inverse of distance

angular momentum depends directly on the root of the distance

Explanation:

1) for this exercise we will use Newton's second law

            F = ma

in this case the acceleration is centripetal

            a = v² / r

the linear and angular variable are related

           v = w r

we substitute

           a = w² r

force is the universal force of attraction

           F = G \frac{m M}{r^2}

we substitute

         G \frac{m M}{r^2} = m w^2 r

         w² = \frac{GM}{r^3}

angular velocity is related to frequency and period

         w = 2π f = 2π / T

we substitute

            ( \frac{2\pi }{T} ) = \frac{GM}{r^3}

the final equation is

             T² = ()  r³

b) the speed of the orbit can be found

           v = w r

            v = \sqrt{\frac{GM}{r^3} } \ r

            v = \sqrt{\frac{GM}{r} }

in this case the dependency is the inverse of the root of the distance

Kinetic energy

           K = ½ M v²

           K = ½ M GM / r

           K = ½ GM² 1 / r

the kinetic energy depends on the inverse of the distance

Potential energy

          U =

          U = -G mM / r

dependency is the inverse of distance

Angular momentum

          L = r x p

for a circular orbit

           L = r p = r Mv

           L =

         L =

The angular momentum depends directly on the root of the distance

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Answer:

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