A rocket is launched vertically such that its acceleration upward is 4.7g. If it can maintain that acceleration, how much time (in minutes) would it take to reach the orbit of the International Space Station 310km above Earth’s surface?
distance = 1/2 * acceleration * time^2
working in meters ... answer in seconds
... 3.1E5 = 1/2 * 4.7 * 9.8 * t^2
find t ... divide by 60 to get minutes
So final answer is 0.07 minutes?
nooo ... more like two
3.1E5 is 310,000
To find the time it takes for the rocket to reach the orbit of the International Space Station, we need to follow these steps:
Step 1: Convert the acceleration from g's to meters per second squared.
The acceleration of 4.7g means that it is 4.7 times the acceleration due to gravity (9.8 m/s^2). So, the acceleration is 4.7 * 9.8 = 45.86 m/s^2.
Step 2: Find the time it takes to reach the desired altitude.
We can use the kinematic equation: s = ut + 0.5at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time. In this case, the displacement is 310 km = 310,000 m, the initial velocity is 0 m/s (since the rocket starts from rest), and the acceleration is 45.86 m/s^2. Plugging these values into the equation, we get:
310,000 = 0 * t + 0.5 * 45.86 * t^2
Rearranging the equation, we get:
0.5 * 45.86 * t^2 = 310,000
Now, let's solve for t.
Step 3: Solve for t.
Divide both sides of the equation by 0.5 * 45.86:
t^2 = 310,000 / (0.5 * 45.86)
t^2 = 12,098
Taking the square root of both sides:
t = √(12,098)
t ≈ 109.98 seconds
Step 4: Convert seconds to minutes.
To convert seconds to minutes, divide the time in seconds by 60:
t (in minutes) = 109.98 / 60 ≈ 1.83 minutes
Therefore, it would take approximately 1.83 minutes for the rocket to reach the orbit of the International Space Station.