Question

. Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 1.00 × 104 kg. The thrust of its engines is 3.00 × 104 N. (a) Calculate the module’s magnitude of acceleration in a vertical takeoff from the Moon. (b) Could it lift off from Earth? If not, why not? If it could, calculate the magnitude of its acceleration.

Answer 1

a) The module's acceleration in a vertical takeoff from the Moon will be
`1.377 (m)/(s^2)`

b) Then we can say that a thrust of
`3*10^(4) N` won't be able to lift off the module from the Earth because it's smaller than the module's weight (
`9.8 *10^(4) N`).

Step-by-step explanation:

a) During a vertical takeoff, the sum of the forces in the vertical axis will be equal to mass times the module's acceleration. In this this case, the thrust of the module's engines and the total module's weight are the only vertical forces. (In the Moon, the module's weight will be equal to its mass times the Moon's gravity acceleration)

`T-(m*g)=m*a`

Where:

`T=` thrust
`=3 *10^(4) N`

`m=` module's mass
`=1 *10^(4) N`

`g=` moon's gravity acceleration
`=1.623 (m)/(s^2)`

`a=` module's acceleration during takeoff

Then, we can find the acceleration like this:

`a=(T)/(m) -g=\frac{3*{10}^4 N}{1*{10}^4 kg}-1.623(m)/(s^2)`

`a=1.377 (m)/(s^2)`

The module's acceleration in a vertical takeoff from the Moon will be
`1.377 (m)/(s^2)`

b) To takeoff, the module's engines must generate a thrust bigger than the module's weight, which will be its mass times the Earth's gravity acceleration.

`weight=m*g=(1*{10}^4 kg)*(9.8 (m)/(s^2))=9.8 *10^(4) N`

Then we can say that a thrust of
`3*10^(4) N` won't be able to lift off the module from the Earth because it's smaller than the module's weight (
`9.8 *10^(4) N`).