Question

Asteroid Ida was photographed by the Galileo spacecraft in 1993, and the photograph revealed that the asteroid has a small moon, which has been named Dactyl. Assume the mass of Ida is 4.4 x 1016 kg, the mass of Dactyl is 2.6 x 1012 kg, and the distance between the center of Dactyl and Ida is 95 km. G = 6.672x10-11 N-m2/kg2 Part Description Answer Save Status A. Assuming a circular orbit, what would be the orbital speed of Dactyl? (include units with

Answer 1

The orbital speed of Dactyl around Ida is approximately 161.6 m/s.

Step-by-step explanation:

The orbital speed of an object can be calculated using the formula v = √ (G * (M + m) / r), where v is the orbital speed, G is the gravitational constant (6.672x10-11 N-m2/kg2), M is the mass of the primary object (Ida), m is the mass of the secondary object (Dactyl), and r is the distance between the center of the two objects.

Plugging in the given values, we have:

v = √ ((6.672 x 10-11 N-m2/kg2) * ((4.4 x 1016 kg) + (2.6 x 1012 kg)) / (95,000 m))

Simplifying the equation, we get:

v = 161.6 m/s

Therefore, the orbital speed of Dactyl around Ida is approximately 161.6 m/s.

Answer 2

The orbital speed of Dactyl is
`5.55m/s`

Step-by-step explanation:

The orbital speed can be determined by the combination of the universal law of gravity and Newton's second law:

`F = G(M \cdot m)/(r^(2))` (1)

Where G is gravitational constant, M is the mass of the asteroid, m is the mass of the moon and r is the distance between them

In the other hand, Newton's second law can be defined as:

`F = ma` (2)

Where m is the mass and a is the acceleration

Then, equation 2 can be replaced in equation 1

`m\cdot a  = G(M \cdot m)/(r^(2))` (2)

However, a will be the centripetal acceleration since the moon Dactyl describe a circular motion around the asteroid

`a = (v^(2))/(r)` (3)

`m(v^(2))/(r) = G(M \cdot m)/(r^(2))` (4)

Therefore, v can be isolated from equation 4:

`m \cdot v^(2) = G (M \cdot m)/(r^(2))r`

`m \cdot v^(2) = G (M \cdot m)/(r)`

`v^(2) = G (M \cdot m)/(rm)`

`v^(2) = G (M)/(r)`

`v = \sqrt{(G M)/(r)}` (5)

Finally, the orbital speed can be found from equation 5:

Notice, that it is necessary to express r in units of meters.

`r = 95km \cdot (1000m)/(1km)` ⇒
`95000m`

`v = \sqrt{((6.672x10^(-11)N.m^(2)/kg^(2))(4.4x10^(16)kg))/(95000m)}`

`v = 5.55m/s`

Hence, the orbital speed of Dactyl is
`5.55m/s`