According to Newton's law of Gravitation, the force $F$ exerted between two bodies of masses $m1$ and $m2$ and separated by a distance $r$ is equal to the product of their masses and inversely proportional to the square of the distance:

$F=G\frac{(m1)(m2)}{r^2}$ (1)

Where $G$ is the gravitational constant

This means that the gravity force decreases when the distance between these two bodies increases.

In this context, if the distance between the capsule and the Earth increases twice, the new distance will be $2r$ .

Substituting this distance in (1):

$F=G\frac{(m1)(m2)}{(2r)^2}$ (2)

$F=G\frac{(m1)(m2)}{4r^2}$

Finally:

$F=\frac{1}{4}G\frac{(m1)(m2)}{r^2}$ >>>This means the force toward Earth becomes one quarter "weaker"

3 0

3 years ago

A converging-diverging nozzle has a throat area of 10 cm2 and an exit area of 28.96 cm2 . A normal shock stands in the exit when

Svetllana [295]

The tank pressure is 5.08 kPa and the mass flow rate is 2.6 kg/s.

The given parameters:

Throat area of the nozzle, $A^*$ = 10 cm² = 0.001 m²
The exit area of the nozzle, A = 28.96 cm² = 0.002896 m²
Air pressure at sea level = 101.325 kPa

The ratio of the areas of the converging-diverging nozzle is calculated as follows;

$= \frac{A}{A^*} \\\\= \frac{0.002896}{0.001} \\\\= 2.896$

From supersonic isentropic table, at $\frac{A}{A^*} = 2.896$ , we can determine the following;

$M_e = 2.6 \ kg/s\\\\\frac{P_o}{P_e} = 19.954$

The tank pressure is calculated as follows;

$\frac{P_o}{P_e} = 19.954 \\\\P_e = \frac{P_o}{19.954} \\\\P_e = \frac{101.325 \ kPa}{19.954} \\\\P_e = 5.08 \ kPa$

Thus, the tank pressure is 5.08 kPa and the mass flow rate is 2.6 kg/s.

Learn more about converging-diverging nozzle design here: brainly.com/question/13889483

8 0

2 years ago