Question

Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its orbital speed (in m/s)? m/s (b) If we want to propel a portion of the rocket to infinity (in the direction tangential to the circular orbit), what's the escape speed from there (in m/s)? m/s

Answer 1

a

The orbital speed is
`v= 2.6*10^(3) m/s`

b

The escape velocity of the rocket is
`v_e= 3.72 *10^3 m/s`

Step-by-step explanation:

Generally angular velocity is mathematically represented as

`w = (2 \pi)/(T)`

Where T is the period which is given as 1.6 days =
`1.6 *24 *60*60 = 138240 sec`

Substituting the value

`w = (2 \pi)/(138240)`

`= 4.54*10^ {-5} rad /sec`

At the point when the rocket is on a circular orbit

The gravitational force = centripetal force and this can be mathematically represented as

`(GMm)/(r^2) = mr w^2`

Where G is the universal gravitational constant with a value
`G = 6.67*10^(-11)`

M is the mass of the earth with a constant value of
`M = 5.98*10^(24)kg`

r is the distance between earth and circular orbit where the rocke is found

Making r the subject

`r = \sqrt[3]{(GM)/(w^2) }`

`= \sqrt[3]{(6.67*10^(-11) * 5.98*10^(24))/((4.45*10^(-5))^2) }`

`= 5.78 *10^7 m`

The orbital speed is represented mathematically as

`v=wr`

Substituting value

`v= (5.78*10^7)(4.54*10^(-5))`

`v= 2.6*10^(3) m/s`

The escape velocity is mathematically represented as

`v_e = \sqrt{(2GM)/(r) }`

Substituting values

`= \sqrt{(2(6.67*10^(-11))(5.98*10^(24)))/(5.78*10^7) }`

`v_e= 3.72 *10^3 m/s`