Question

Two satellites, X and Y, are orbiting Earth. Satellite X is 1.2 × 106 m from Earth, and Satellite Y is 1.9 × 105 m from Earth. Which best compares the satellites? Satellite X has a greater period and a faster tangential speed than Satellite Y. Satellite X has a greater period and a slower tangential speed than Satellite Y. Satellite X has a shorter period and a faster tangential speed than Satellite Y. Satellite X has a shorter period and a slower tangential speed than Satellite Y.

Answer 1

B

Step-by-step explanation:

Answer 2

Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

Step-by-step explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

`T^(2)=(4\pi^(2))/(GM)r^(3)` (1)

Where;

`G=6.674(10)^(-11)(m^(3))/(kgs^(2))` is the Gravitational Constant

`M=5.972(10)^(24)kg` is the mass of the Earth

`r` is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

So for satellite X, the orbital period
`T_(X)` is:

`T_(X)^(2)=(4\pi^(2))/(GM)r_(X)^(3)` (2)

Where
`r_(X)=1.2(10)^(6)m`

`T_(X)^(2)=(4\pi^(2))/((6.674(10)^(-11)(m^(3))/(kgs^(2)))(5.972(10)^(24)kg))(1.2(10)^(6)m)^(3)` (3)

`T_(X)=413.712 s` (4)

For satellite Y, the orbital period
`T_(Y)` is:

`T_(Y)^(2)=(4\pi^(2))/(GM)r_(Y)^(3)` (5)

Where
`r_(Y)=1.9(10)^(5)m`

`T_(Y)^(2)=(4\pi^(2))/((6.674(10)^(-11)(m^(3))/(kgs^(2)))(5.972(10)^(24)kg))(1.9(10)^(5)m)^(3)` (6)

`T_(Y)=26.064 s` (7)

This means
`T_(X)>T_(Y)`

Now let's calculate the tangential speed for both satellites:

For Satellite X:

`V_(X)=\sqrt{(GM)/(r_(X))}` (8)

`V_(X)=\sqrt{((6.674(10)^(-11)(m^(3))/(kgs^(2)))(5.972(10)^(24)kg))/(1.2(10)^(6)m)}`

`V_(X)=18224.783 m/s` (9)

For Satellite Y:

`V_(Y)=\sqrt{(GM)/(r_(Y))}` (10)

`V_(Y)=\sqrt{((6.674(10)^(-11)(m^(3))/(kgs^(2)))(5.972(10)^(24)kg))/(1.9(10)^(5)m)}`

`V_(Y)= 45801.13 m/s` (11)

This means
`V_(Y)>V_(X)`

Therefore:

Satellite X has a greater period and a slower tangential speed than Satellite Y