Evan's Space

Wonders of Physics


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Fiery Re-entry into Earth’s Atmosphere (updated)

This post was updated following the first astronauts launched by SpaceX returned home safely on 3 Aug 2020.

SpaceX’s Crew Dragon heat shield shown off after first orbital-velocity reentry

How do spacecraft re-enter the Earth? | HowStuffWorks

Why Is It So Difficult For A Returning Spacecraft To Re-Enter Our Atmosphere?

Returning from Space: Re-entry – PDF format

SpaceX In-Flight Abort Test

SpaceX Falcon Heavy- Elon Musk’s Engineering Masterpiece

Shuttle Atlantis STS-132 – Amazing Shuttle Launch Experience

How to Land the Space Shuttle… from Space

First astronauts launched by SpaceX return to earth (3 Aug 2020)


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Understanding Newton’s 1st and 2nd Laws of Motion

1) Newton’s first law states that an object will remain at rest or in uniform motion (constant speed) in a straight line unless an external force acts on the body.

In other words, when a body is at rest or moving at constant speed in a straight line (constant velocity), straight away you should know it is Newton’s first law. Next you must know these 3 basics concepts about 1st law:
– forces acting on the body are balanced
– net force / resultant force acting on the body is zero
– there is no acceleration.

2) Newton’s second law states that when a net force (resultant force) acts on a body, it will cause an acceleration on the body (accelerating or decelerating).

Basically F = ma where F is the net or resultant force in N,
m is the mass in kg
a is the acceleration in ms-2

In other words, when a body is moving faster or slower (or going round a bend), you should know its Newton’s second law. Next you must know these 3 basic concepts about 2nd law:
– forces acting on the body are not balanced
– there is a net force / resultant force acting on the body
– there is an acceleration
(accelerating of net force is in the direction of motion, or decelerating if the net force is opposite to the direction of motion)

Click here to know more about Newton’s 3rd Law


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Parachute Jump – Speed-time graph

In a typical parachute jump, there are various distinct stages/sections as you can see from the graph in the video below.

AB = constant acceleration, free fall, a = 10 ms-2
BC = decreasing acceleration
CD = constant speed, zero acceleration
at D = the time where the parachute is fully opened
DE = constant deceleration
EF = lower constant speed, zero acceleration
FG = constant deceleration

After you have learned Dynamics, you should be able to explain each stage using forces acting on the skydiver, namely the weight and air resistance.

Refer to the video below to understand the motion at various stages and how to explain in terms of forces, esp the part on why the acceleration is decreasing during BC.

You can refer to the detailed explanation in words in the comics below. Hope this post helps you to understand better!

velocity-time graph of parachute jump


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Impressive Bugatti Chiron

This impressive Bugatti Chiron can accelerate from rest to 400 km/h and decelerate to a complete stop in merely 42 seconds! Our normal cars on the expressway travel about 90 km/h and the F1 race yesterday night is about 300 km/h. This Bugatti Chiron is faster than most bullet trains and comparable to the speed of a magnetic levitation train!

Capturehttps://www.bugatti.com/chiron

Before we look at the video, let’s do some calculations:

Capture2

Let’s find the acceleration of the car to reach 400 km/h in 32.6 sec:

Converting 400 km/h to m/s:    400km/1h = 400 000m/3600s =111 m/s

acceleration, a = (v – u)/t = (111 – 0) / 32.6 = 3.4 m/s2

hmmm…. this acceleration doesn’t seem impressive… it is way below free fall acceleration!

But we are not being fair here. To achieve the max speed of 400 km/h is not easy due to the resistive force (air resistance and friction) as speed increases. We should compare fairly the acceleration to reach 100 km/h instead like how we typically compare sports car like Ferrari etc.

Let’s find the acceleration of the car to reach 100 km/h (27.8 m/s) in 2.4 sec:

acceleration, a = (v – u)/t = (27.8 – 0) / 2.4 = 11.6 m/s2

This is greater than acceleration due to gravity (free fall) and much faster than most sports cars in the market like Ferrari or Lamborghini!

Now, let’s find the deceleration of the car when it slows down from 400 km/h to a complete stop in 41.9 – 32.6 = 9.36 s

acceleration, a = (v – u)/t = (0 – 111) / 9.36 = -11.9 m/s2

Take note of the spoiler being activated when it decelerates. This increases the drag (air resistance) to slow down the car, in addition to using the normal brakes. It is the same principle as the aeroplane when it lands and slows down on the runway.