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Anna35 [415]
3 years ago
5

Two satellites A and B orbit the Earth in the same plane. Their masses are 5 m and 7 m, respectively, and their radii 4 r and 7

r, respectively What is the ratio of their orbital speeds
Physics
1 answer:
Dmitry [639]3 years ago
8 0

Answer:

The ratio of their orbital speeds are 5:4.

Explanation:

Given that,

Mass of A = 5 m

Mass of B = 7 m

Radius of A = 4 r

Radius of B = 7 r

The orbital speed of satellite A,

v_{A}=\sqrt{\dfrac{GM_{A}}{R_{A}}}......(I)

The orbital speed of satellite B,

v_{B}=\sqrt{\dfrac{GM_{B}}{R_{B}}}......(I)

We need to calculate the ratio of their orbital speeds

Using equation (I) and (II)

\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{\dfrac{GM_{A}}{R_{A}}}{\dfrac{GM_{B}}{R_{B}}}}

Put the value into the formula

\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{G\times5m\times7r}{G\times7m\times4r}}

\dfrac{v_{A}}{v_{B}}=\dfrac{5}{4}

Hence, The ratio of their orbital speeds are 5:4.

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A 5-kg ball collides inelastically head-on with a 10-kg ball, which is initially stationary. Which of the following statements i
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Answer:

The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball.

Explanation:

In inelastic collision, the total momentum is always conserved after collision but the kinetic energy is reduced.

Momentum is Mass X velocity.

5 kg ball is in motion, while 10 kg ball is stationary; that is zero velocity.

The momentum of 10 kg ball before collision is zero while the momentum of 5 kg ball before collision is more than zero. Therefore, the magnitude of change in momentum will not be equal.

Next possible options are in kinetic Energy

Initial Kinetic energy = \frac{1}{2}mu^2

Final kinetic energy =\frac{1}{2}mv^2

Change in kinetic energy = Final Kinetic Energy - Initial Kinetic Energy

Change in kinetic energy of 5kg ball = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2

Since the 5-kg ball has initial velocity (u), the magnitude of the change in velocity will be reduced.

Change in kinetic energy of 10kg ball:

the ball is initially at rest, therefore the initial velocity (u) will be zero (0)

Δ K.E = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2 = \frac{1}{2}m(v-0)^2 = \frac{1}{2}mv^2

From the solution above, the magnitude of the change in velocity experienced by 10 kg ball is higher than 5 kg ball.

Hence, The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball

4 0
3 years ago
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