what are the forces of a rocket taking off?

Answer 1

A rocket is subjected to four forces in flight; weight, thrust, and the aerodynamic forces, lift and drag.

at rest I will denote the launch pressure by the symbol ps. The pressure exerts an equal force on all sides of the pressure vessel, the cone of the engine. Because the rocket is free to move along the launch mechisims, the pressure exerts a vertical force given by:

Fp = (ps - po) * A

where Fp is the pressure force, po is the atmospheric pressure, and A is the cross-sectional area of the rocket tube. The net force acting on the rocket is the difference of the vertical pressure force and the weight W of the rocket:

Fnet = (ps - po) * A - W

To simplify the analysis, we are going to assume that the length of the launch tube is small relative to the length of the rocket. As the rocket moves, the pressure remains fairly constant inside the tube. In reality, there is a slight decrease in pressure as the volume increase, but we are going to neglect that effect to simplify this analysis. With a constant pressure, and constant pressure force, we can use a simple algebraic form of Newton's second law to determine the acceleration a, velocity v and the distance y which the rocket moves along the launch tube:

Fnet = m * a

m * a = (ps - po) * A - W

where m is the mass of the rocket This equation can also be written as:

W * a = g * ((ps - po) * A - W)

where g is the gravitational acceleration, which is equal to 32.2 ft/sec^2 or 9.8 m/sec^2 on the surface of the Earth. The value of g is different on the Moon, and on Mars. The weight W is equal to the mass times the gravitational acceleration. Now divide both sides by the weight to determine the rocket acceleration along the launch tube:

a = g * ((ps - po) * A / W - 1)

For a constant acceleration, the velocity v and distance traveled y are given by:

v = a * t

y = .5 * a * t^2

where t is the time. Using the length of the launch tube L, we can now solve for the lift off time - TLO by using the second equation:

TLO = sqrt (2 * L / a)

and then using this time, we can use the first equation to solve for the velocity at the end of the tube:

v = TLO * a

v = sqrt ( 2 * L * a)

As the rocket clears the end of the launch tube (t = TLO+), the air inside the rocket, at pressure ps, will rush out the exit and the pressure will equalize with free stream inside the tube. In reality, this will provide an additional amount of thrust T given by:

T = m dot * Ve

where m dot is the air mass flow rate out of the tube, Ve is the exit velocity of the air. For an air rocket, with no nozzle and small volume, this additional thrust is small relative to the weight of the rocket, and we will ignore this effect.

The velocity v at the end of the launch tube is used as the initial velocity in the flight equations to determine the flight trajectory of a rocket.

Answer 2

Thrust. Acceleration of mass in a unmarried course (ie out of the exhaust of a rocket) causes the article to be moved contained in the alternative course with equivalent rigidity (see Newton's guidelines.) the quicker the speed, the further thrust is generated. power = thrust x p.c.. The gravitational acceleration is expressed contained in the load-to-thrust-ratio. A rocket could generate adequate thrust to achieve and preserve a minimum of 11.2km/s to flee the Earth's gravity.

Answer 3

Mostly thrust and weight. After it gets up to speed but before it exits the atmosphere there is drag too. That is what NASA calls max Q or maximum dynamic pressure, where increasing speed and decreasing air pressure are balanced to cause maximum drag. They throttle the engines back a little at that point to reduce stress on the vehicle. After it gets high enough for the air to be so thin as to not be a problem, they throttle the engines back up to full power. That is what they mean by, "go for throttle up".

Answer 4

The thrust (F) of the rocket engine must be sufficient to overcome gravity (g) and also to accelerate the rocket (mass = m) into the air (with acceleration = a).

So F = m(g+a)

Answer 5

They have to get to 17,500 mph to escape from the earths gravity that's called the escape velocity