According to the minimum thrust to weight ratio of 4:1, what is the maximum mass a motor with 65 Ns can safely lift?

To determine the maximum mass that can be safely lifted by a motor with a specific impulse of 65 Newton-seconds (Ns), while adhering to the minimum thrust-to-weight ratio of 4:1, we can use the following principles.

The thrust-to-weight ratio is defined as the ratio of the thrust produced by the rocket motor to the weight of the rocket. To achieve a thrust-to-weight ratio of 4:1, the thrust must be four times greater than the weight of the rocket.

Using the formula for weight (W = m * g), where g is the acceleration due to gravity (approximately 9.81 m/s²), we can express weight in terms of mass:

Weight = m * g

For a thrust-to-weight ratio of 4:1, it follows that:

Thrust ≥ 4 * Weight

Thrust ≥ 4 * (m * g)

Given that the motor produces a thrust equal to its impulse output, we know that:

Thrust = 65 Ns

Substituting this into the inequality, we get:

65 Ns ≥ 4 * (m * 9.81 m/s²)

Now, rearranging the inequality to find the maximum mass (m):

m ≤ 65