A rocket is fired in deep space, where gravity is negligible. In the first second it ejects 1/160 of its mass as exhaust gas and has an acceleration of 15.1 m/s^2.What is the speed v_gas of the exhaust gas relative to the rocket?
To find the speed of the exhaust gas relative to the rocket, we can use the principle of conservation of momentum.
The momentum of the rocket-gas system must remain constant in the absence of external forces. Initially, the rocket is at rest, so the momentum is zero. After ejecting the exhaust gas, the rocket will have a non-zero momentum, which means the exhaust gas must also have an equal and opposite momentum.
Given:
Mass of the rocket = m
Mass of the exhaust gas = 1/160 of m, which is (1/160)m
Acceleration of the rocket = 15.1 m/s^2
Let's determine the speed of the exhaust gas relative to the rocket (v_gas) using the conservation of momentum:
Initial momentum = Final momentum
(0) = (m + (1/160)m) * v_gas
Since gas is ejected in one second, its velocity relative to the rocket can be considered constant. Therefore, we can write:
0 = (161/160)m * v_gas
Simplifying this equation, we find:
0 = (161/160)v_gas
To solve for v_gas, set the equation equal to zero:
(161/160)v_gas = 0
Divide both sides of the equation by (161/160):
v_gas = 0
Therefore, the speed of the exhaust gas relative to the rocket is zero.
To find the speed of the exhaust gas relative to the rocket, we need to use the principle of conservation of momentum. The total momentum of the system (rocket + exhaust gas) is conserved.
Let's assume the initial mass of the rocket is m and the mass of the exhaust gas ejected in the first second is Δm (1/160 of the rocket's mass).
The initial momentum of the rocket is zero since it is initially at rest in deep space. So, the total momentum after the ejection of the exhaust gas should also be zero.
The momentum of the rocket after the ejection is given by its mass multiplied by its velocity, which we can denote as V_rocket. The momentum of the exhaust gas is given by its mass multiplied by its velocity, denoted as V_gas.
Using the principle of conservation of momentum, we can write the equation:
(m - Δm) * V_rocket + Δm * V_gas = 0
We are given the acceleration of the rocket, which is 15.1 m/s^2. Using the equation a = ΔV/Δt, where a is the acceleration, ΔV is the change in velocity, and Δt is the time interval, we can determine ΔV for the first second.
15.1 m/s^2 = ΔV / 1s
ΔV = 15.1 m/s
Now, let's express V_rocket and V_gas using this ΔV and the given information that the exhaust gas ejected is 1/160 of the rocket's mass.
V_rocket = -ΔV
V_gas = v_gas - (-ΔV) [since velocity is relative]
Substituting these values into the conservation of momentum equation:
(m - Δm) * (-ΔV) + Δm * (v_gas - (-ΔV)) = 0
Simplifying the equation:
(-m + Δm) * ΔV + Δm * (v_gas + ΔV) = 0
Since the rocket ejects 1/160 of its mass as exhaust gas, Δm = m/160.
(-m + (m/160)) * ΔV + (m/160) * (v_gas + ΔV) = 0
Multiplying through by 160 to remove the denominators:
(-160m + m) * ΔV + m * (v_gas + ΔV) = 0
Simplifying further:
(-159m) * ΔV + m * (v_gas + ΔV) = 0
Dividing through by m to isolate v_gas:
(-159) * ΔV + v_gas + ΔV = 0
Rearranging the equation:
v_gas = 159 * ΔV - ΔV
Substituting the value of ΔV we found earlier:
v_gas = 159 * 15.1 m/s - 15.1 m/s
Calculating the value:
v_gas = 2403.9 m/s - 15.1 m/s
v_gas = 2388.8 m/s
Therefore, the speed of the exhaust gas relative to the rocket is approximately 2388.8 m/s.
A rocket is fired in deep space, where gravity is negligible. In the first second it ejects 1/160 of its mass as exhaust gas and has an acceleration of 15.1 m/s^2.What is the speed v_gas of the exhaust gas relative to the rocket?
By means of the rocket equation,
deltaV = cln(Wo/Wt) where
deltaV or dV = the incremental change in velocity over a given time period, Wo = the initial weight at ignition, Wt = the weight after a specific time period and c = the velocity of the exhaust gases. ln = the natural logarithm of.
After one second of firing, the rocket achieves a speed of 15.m/s.
Therefore,
dV = 15.1 = cln[Wo/(Wo-Wo/160)
...= ]5.1 = cln[1/(1-1/160] = ............cln(1.006289)
15.1 = cln(1.006289) - c(.0062696
Hence, c = 2408m/s