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Alex
3 years ago
15

Which statement is true about a planet’s orbital motion?

Physics
1 answer:
lana66690 [7]3 years ago
3 0

Answer:

Orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.

Explanation:

The gravitational force is responsible for the orbital motion of the planet, satellite, artificial satellite, and other heavenly bodies in outer space.

When an object is applied with a velocity that is equal to the velocity of the orbit at that location, the body continues to move forward. And, this motion is balanced by the gravitational pull of the second object.

The orbiting body experience a centripetal force that is equal to the gravitational force of the second object towards the body.

The velocity of the orbit is given by the relation,

                                    V = \sqrt{\frac{GM}{R + h} }

Where

                   V - velocity of the orbit at a height h from the surface

                    R - Radius of the second object

                    G - Gravitational constant

                    h - height from the surface

The body will be in orbital motion when its kinetic motion is balanced by gravitational force.

                         1/2 mV^{2} = GMm/R

Hence, the orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.

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expeople1 [14]

Answer:

The final velocity of the cars is 8.38 m/s at an angle 39.1° south of west.

Explanation:

Given that,

Mass of first car = 1100 kg

Velocity of first car = 8.5 m/s

Mass of second car = 650 kg

Velocity of second car = 17.5 m/s

Suppose we need to find the final velocity of the cars and direction of the cars.

We need to calculate the velocity of the car in west direction

Using conservation of momentum in west direction

m_{f}v_{f}+m_{s}v_{s}= (m_{f}+m_{s})v_{x}

v_{x}=\dfrac{m_{f}v_{f}+m_{s}v_{s}}{(m_{f}+m_{s})}

Put the value into the formula

v_{x}=\dfrac{1100\times0+650\times17.5}{1100+650}

v_{x}=6.5\ m/s

We need to calculate the velocity of the car in south direction

Using conservation of momentum in south direction

m_{f}v_{f}+m_{s}v_{s}= (m_{f}+m_{s})v_{y}

v_{y}=\dfrac{m_{f}v_{f}+m_{s}v_{s}}{(m_{f}+m_{s})}

Put the value into the formula

v_{y}=\dfrac{1100\times8.5+650\times0}{1100+650}

v_{y}=5.3\ m/s

We need to calculate the final velocity of the cars

Using formula of velocity

v_{eq}=\sqrt{(6.5)^2+(5.3)^2}

v_{eq}=8.38\ m/s

We need to calculate the direction

Using formula of direction

\tan\theta=\dfrac{v_{y}}{v_{x}}

Put the value into the formula

\tan\theta=\dfrac{5.3}{6.5}

\theta=\tan^{-1}(\dfrac{5.3}{6.5})

\theta=39.1^{\circ}

Hence, The final velocity of the cars is 8.38 m/s at an angle 39.1° south of west.

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Answer:

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Explanation:

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2 years ago
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sergij07 [2.7K]

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Explanation:

KE = (1/2)mv^2

103kJ = 103000J

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