Question

Rank the following objects, in order of their circular velocities, from smallest to largest

a. a 5-kg object orbiting Earth halfway to the moon

b. a 10-kg object orbiting Earth just above Earth's surface

c. a 15-kg object orbiting Earth at the same distance as the Moon

PLEASE SHOW DETAIL ON CALCULATIONS

Answer 1

Answer:
`V_(1)<V_(3)<V_(2)`

a)
`V_(1)=1.317(10)^(-9) m/s`

b)
`V_(2)=1.02(10)^(-8) m/s`

c)
`V_(3)=1.61(10)^(-9) m/s`

Explanation:

Assuming these objects are moving with uniform circular motion (and the orbit is perfectly circular), the velocity will be given by the following equation:

`V=\sqrt{G(m)/(r)}`

Where:

`V` is the velocity of the object, which is assumed as constant.

`G=6.67408(10)^(-11)(m^(3))/(kgs^(2))` is the Gravitational Constant

`m` is the mass of the object

`r` is the radius of the orbit

Knowing this, let's begin with the answers:

a. a 5-kg object orbiting Earth halfway to the moon

In this case the radius of the orbit is the half of the distance between the Earth and the moon:

`r_(1)=(d_(Earth-Moon))/(2)=(384400000 m)/(2)`

`r_(1)=192200000 m`

And the mass will be
`m=5 kg`. So, the equation of the velocity is:

`V_(1)=\sqrt{G(m_(1))/(r_(1))}`

`V_(1)=\sqrt{6.67408(10)^(-11)(m^(3))/(kgs^(2))(5 kg)/(192200000 m)}`

`V_(1)=0.000000001317 m/s=1.317(10)^(-9) m/s` This is the velocity of the first object

b. a 10-kg object orbiting Earth just above Earth's surface

Now the radius of the orbit is approximately the radius of the Earth:

`r_(2)=r_(Earth)=6371000 m`

And the mass is
`m_(2)=10 kg`. So, the equation of the velocity is:

`V_(2)=\sqrt{G(m_(2))/(r_(2))}`

`V_(2)=\sqrt{6.67408(10)^(-11)(m^(3))/(kgs^(2))(10 kg)/(6371000 m)}`

`V_(2)=0.0000000102 m/s=1.02(10)^(-8) m/s` This is the velocity of the second object

c. a 15-kg object orbiting Earth at the same distance as the Moon

With this last case, the radius of the orbit is equal to the distance between the Earth and the Moon:

`r_(3)=r_(Earth-Moon)=384400000 m`

And the mass is
`m_(3)=15 kg`. So, the equation of the velocity is:

`V_(3)=\sqrt{G(m_(3))/(r_(3))}`

`V_(3)=\sqrt{6.67408(10)^(-11)(m^(3))/(kgs^(2))(15 kg)/(384400000 m)}`

`V_(3)=0.00000000161 m/s=1.61(10)^(-9) m/s` This is the velocity of the third object

Finally, comparing the velocity of the three objects we have:

`1.317(10)^(-9) m/s < 1.61(10)^(-9) m/s < 1.02(10)^(-8) m/s`

`V_(1) <V_(3) < V_(2)`