Morning n' Morning n' Morning!

You are in a jet plane flying along the Equator, wearing a watch that uses GPS location to update the time.

Right now, it's 8am in the morning.
One hour later, it's still 8am in the morning.
Five hours later, it's still 8am in the morning!

There is a maximum, and a minimum speed at which this feat can be accomplished. Find v max + v min 2 \frac{v_\text{max} + v_\text{min}}{2} in m/s, of course!

Assumptions and Details

  • The Earth is a perfect sphere with radius 6370 km
  • Your plane has a constant ground speed.


The answer is 463.23947.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Jessica Wang
Oct 26, 2015

v = ω r = 2 π r T = 2 π × 6370 × 1 0 3 24 × 60 × 60 = 463.24 m / s v=\omega r = \frac{2\pi r}{T}=\frac{2\pi \times 6370\times 10^{3}}{24\times 60\times 60}= \boxed{463.24}m/s , to 2d.p.

Moderator note:

Since local time zones are wide (and not extremely local), there is actually a whole range of times in which this could work. For example, we could be travelling at a speed of 1 km / hour, and still cross a time zone to make it still 8am.

Nice question, and a beautiful solution. +1!

Sravanth C. - 5 years, 7 months ago

@Vijay Simha @Arifur Rahman Just fixed -- well spotted :)

Jessica Wang - 5 years, 7 months ago

Good question

Sangamesh Rk - 5 years, 7 months ago

Log in to reply

You can give it a try - Sravanth Chebrolu's Gaining Time .

Muhammad Arifur Rahman - 5 years, 7 months ago

You need to specify the units. 463.24 What ?

Vijay Simha - 5 years, 7 months ago

Log in to reply

Yeah, most probably she forgot. Wait, she will be fixing the typo ASAP.

Muhammad Arifur Rahman - 5 years, 7 months ago

Am I the only person that thinks that the unit that an answer is given in should be specified in the problem details and assumptions. I think this is a very good problem I suppose it could be argued that m/s is the standard unit of speed used form most calculations especially at slower speed rates. I mistakenly assumed the answer would be given in a much more common unit of speed used for aircraft ie km/hr. I think the details and assumptions of this problem should be edited to indicate m/s are expected in the solution.

Also in practical terms thinking about the maximum speed possible in a jet aircraft this is a correct solution. However theoretically there is no real maximum speed that the aircraft could be traveling at to make this possible. The limit would be set by the limit of the maximum speed of the jet plane. If we do away with the practical limitations set by the speed possible with a modern jet plane there is no upper limit to the speed that the plane could be traveling. The maximum speed could be unlimited if it was possible to achieve speeds approaching or exceeding one revolution around the earth each hour.

1/24 earth revolution per hour west or 23/24 revolution per hour east. 1+1/24 or 2+1/24 , n+1/24 west, n+23/24 east revolution per hour would all satisfy the conditions of this problem, the upper limit is set only by the maximum speed of the plane.

Darryl Dennis - 4 years, 11 months ago

What if the plane travels in the opposite direction, fast enough to cover 23/24 of the circumference of earth in 1 hour?

Pulkit Sinha - 5 years, 6 months ago

The earth spins at 463.24 m/s appox.And if the plane is to stay at the same position throught out it's entire journey, then it must travel at 463.24 m/s in the opposite direction. However the answer required is to be expressed in m/s "relative to Earth's surface". Therefore shouldn't we revert and add earths velocity to the plane, which would bring up the answer to 926.48 m/s

Ritesh Singh - 5 years, 7 months ago

Log in to reply

That's a great question! Let me see if I can help explain it. I like to think of this scenario like a treadmill or a conveyor belt. Everything depends on your frame of reference . Let's suppose that I have a conveyor belt that is moving at 10 m/sec . If I am standing "still" on that conveyor belt, then relative to the conveyor belt , my velocity is 0 m/sec . However, relative to my surroundings (the ground, other people, trees, buildings, etc.), my velocity is 10 m/sec in the direction that the conveyor belt is moving.

Similarly, if I am on the conveyor belt and walk at 10 m/sec either with or against its direction of movement, then relative to the conveyor belt my velocity is 10 m/sec . However, relative to my surroundings my velocity is 0 m/sec if I am walking against its direction of movement, and 20 m/sec if I am walking with its direction of movement. (This is because we get my velocity plus the conveyor belt's velocity. Note that while this is true for smaller velocities, it is slightly different for higher velocities. But that's not the point of this example. Click here for more information about it.)

Now apply this to our jet and the Earth's surface. If the jet is traveling at 463.24 m/sec opposite the rotation of the Earth on its axis, then relative to the Earth's surface , the jet's velocity is 463.24 m/sec to the West . However, relative to the surroundings (i.e. our Solar System), the jet has a velocity of 0 m/sec , which is why the time remains the same.

Nate Clegg - 5 years, 7 months ago

Log in to reply

Thanks for the explanation Nate. I must admit that I had some trouble deciding my frame of reference. I imagined myself as a point on the earth's surface which led to the conclusion that I must have a velocity of 463 m/s due east. However I should have considered the earth's surface as a whole instead of considering a mere point. The GPS thingy led me astray for a while but Nonetheless, It was a great problem.

Ritesh Singh - 5 years, 7 months ago

Log in to reply

@Ritesh Singh You can create hundreds of problems from this concept.

Can you generate a similar problem with extraordinary calculations and Post it?

Muhammad Arifur Rahman - 5 years, 7 months ago

Log in to reply

@Muhammad Arifur Rahman Well I will try my best.Although, I am not quite sure about that extraordinary calculations part :3

Ritesh Singh - 5 years, 7 months ago

Log in to reply

@Ritesh Singh Be at Brilliant Lounge and message me. We can clarify that!

Muhammad Arifur Rahman - 5 years, 7 months ago

You ignored the height of a jet plane, anyway the question was incomplete

Aditya Paul - 5 years, 7 months ago
Nate Clegg
Oct 27, 2015

To start off, I must first say that I often find that the solutions provided tend not to fully explain the process by which the answer was found. I've found that I best understand the concept being presented if I have the process explained to me step-by-step . For this purpose, my solution is lengthy and detailed , explaining the math and reasoning behind each step in the process. If you are like me and need step-by-step details , I hope you find my solution to be beneficial.

I thought of this scenario like a treadmill. In order for you to remain in the same place relative to your surroundings (the light fixtures, furniture, walls, etc.), you must be running at the same speed that the treadmill's track is moving . (Note that you must be running opposite the direction that the treadmill's track is moving.)

Time of day is basically a representation of the Sun's position in the sky, relative to where you are on the ground. That positioning changes as the Earth rotates, thus changing the portion of the Earth that receives sunlight. For the Sun to be in the same position relative to the jet (and thus the time remain the same relative to the jet), the jet must travel at the same velocity as the Earth's rotation. So, to solve this problem we must find the velocity of the Earth's rotation in meters per second .

We know that velocity is equal to distance per time: v = d t v = \frac{d}{t} Let our distance d be the circumference of the Earth, d = 2 π r d = 2\pi r where r is the radius of the Earth, which is 6370 km . Let our time t be the time it takes the Earth to complete one full rotation, which is 24 hours . Note that we could use any distance on the Earth's surface and the time it takes to travel that distance, but the circumference and 24 hours are known values and are easy to work with. We will use them for simplicity.

Now that we know our variables, it's time to set up the problem.

We will first replace d in our velocity equation with our equation for the circumference of the Earth. v = 2 π r t v = \frac{2\pi r}{t} Now let's replace the r and t with our radius and time variables, respectively. (Make sure to pay attention to the units , because they're important!) v = 2 π × 6370 k m 24 h r s v = \frac{2\pi\times6370km}{24hrs} We'll go ahead and simplify a little bit by reducing the fraction to make things cleaner. (Not necessary, but it helps me keep track of things more easily.) v = π × 6370 k m 12 h r s v = \frac{\pi\times6370km}{12hrs} Now we have a problem. Our units are in km per hour , but we need them to be in meters per second . To fix this, we need to convert our units . We know that 1 km = 1000 m, 1 hr = 60 min, and 1 min = 60 seconds . Now if we substitute these values into our equation, we get v = π × 6370 × 1000 m 12 × 60 m i n = π × 6370 × 1000 m 12 × 60 × 60 s e c v = \frac{\pi\times6370\times1000m}{12\times60min} = \frac{\pi\times6370\times1000m}{12\times60\times60sec} Now that we have our units right, we can go ahead and use our calculator to solve our equation and find the answer! v = π × 6370 × 1000 m 12 × 60 × 60 s e c = 463.24 m s e c v = \frac{\pi\times6370\times1000m}{12\times60\times60sec} = \boxed{463.24\frac{m}{sec}} Also note that the jet must be traveling west , which is opposite the direction of the Earth's rotation.

463 would assume it is travelling at ground level which should really be stated.

Lawrence Mayne - 5 years, 7 months ago

Log in to reply

I did a little research and found that commercial jets generally fly at altitudes between 33,000 ft (roughly 10 km) and 39,000 ft (roughly 12 km).

So if we added 10 km to our radius, we would get 463.97 m s e c \boxed{463.97 \frac{m}{sec}} and if we added 12 km we would get 464.11 m s e c \boxed{464.11 \frac{m}{sec}} I know that this problem has some tolerance in the answer, because I simply entered 460 m s e c \boxed{460 \frac{m}{sec}} as my original answer, and it was accepted just fine. But you're right, the plane's altitude does slightly change the necessary velocity, simply because it makes the circumference that the jet has to travel slightly larger!

Nate Clegg - 5 years, 7 months ago

Log in to reply

As we haven't taken that a consideration, we can neglect that. Because the problem has already been jumbled enough.

BTW, you pointed a good thing.

Muhammad Arifur Rahman - 5 years, 7 months ago

We've made the modification.

Muhammad Arifur Rahman - 5 years, 7 months ago

Good point. Thanks for bringing that up.

Nate Clegg - 5 years, 7 months ago

The best ever solution for a Level 3 problem.

Greetings.

Muhammad Arifur Rahman - 5 years, 7 months ago
Achille 'Gilles'
Oct 27, 2015

Have you wrote the whole solution with LaTeX ? The solution text looks so BLURRED.

Please use LaTeX for math input ONLY.

BTW, good, precise solution. UPVOTE!

Muhammad Arifur Rahman - 5 years, 7 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...