Tag Info

New answers tagged

1

Start digging at the equator and move all the dirt to the polar regions. This will decrease the moment of inertia of the planet about its spinning axis. Due to the conservation of angular momentum this will result in an increase in angular velocity, akin to a figure skater who retracts her arms while spinning.


3

Cover it in mirrors that are highly reflective on one side and painted black on the other. Position the mirrors so that the "faces" are perpendicular to the surface. A sketch is below (I have only shown three mirrors, the idea is that you would cover the planet with them, but they will be most effectively placed close to the equator). The plan is that each ...


0

As a first step you can use a simple method, which is for every small time step $\Delta t$ approximate the acceleration as constant and use $\Delta v=a\Delta t$ for each direction, and then $\Delta x = v\Delta t$. These equations apply separately for each dimension, so calculate the x and y velocities first and then the resulting changes in position


1

You asked a similar question on worldbuilding for a story. Hit youtube for Operation Crossroads, Baker test. The US Navy detonated a 21kt device 90 feet underwater. Produced a nice fountain but no overly-destructive wave action after a few kilometers. A few years later, Castle Bravo (15Mt) also failed to produce any significant damage* outside the ...


0

Based on the answers here is my summary: Kinetic energy of one plane: $$ E_{kin} = 2 * 10^9 J = 0.5 \text{ kilotons of TNT} $$ Chemical energy of the plane fuel (very close to the amazing estimation of Floris): $$ E_{chem} = 38000 L * 35 * 10^6 J/L = 1.33*10^{12} J = 330 \text{ kilotons of TNT} $$ Potential energy of one collapsing tower: $$ E_{pot} = ...


4

Assuming kerosene is C8H18, has 25 chemical bonds, each of which releases 1eV when burned, gives an energy in fuel of 20 MJ/kg. The weight of a plane shortly after take off is significantly fuel. If I'd guessed I would have said 20 tonnes per plane; (Floris' comment suggests more like 30). This gives 400 GJ per plane, equivalent to 100 tons of TNT per ...


4

Estimating is always a fun aspect of physics - so let's do some, without looking up any values. What is the kinetic energy of a plane? We need to know the mass of a plane and its speed. I am going to use seriously rounded numbers - let's see how close we get. We "know" a full size car is about 1000 kg, and can carry 5 passengers of 100 kg. That means a car ...


-1

Aircraft fuel is kerosense. Sometimes it is difficult to directly measure the amount of heat something produces. We can make the process easier by burning an amount of the fuel to heat water. The energy lost by the fuel can then be calculated by finding the heat gained by the water as measured by the change in temperature of the water. this reference ...


2

One favorite calculation method for approximating diffusion (heat, mass, whatever) is to use the following pretty simple relation: $$L^2 = C D t_c$$ in which you've got length (radius in this case), a constant dependent on geometry (6 for a sphere), diffusivity, and characteristic time. Here you know all but $t_c$, for which you can easily solve (~21s). ...


1

Since the object is small and spherical, and the temperature difference is small, I am going to assume there is no convection - meaning that you basically are asking about a conduction problem. Thermal conductivity of air is 0.0257 W/m/K at 20 °C, and heat capacity is 1.005 kJ/kg/K Details of the calculation (including an animation) are nicely shown in ...


0

I've read a little bit on the science of a space elevator and it's a surprisingly difficult problem. To have a working space elevator, it would need to be at least to the Geosynchronous orbit, 22,000 miles up, probobly a bit beyond that for buoyancy. The highest balloon is some 25 miles - so that's less than 1/10th of 1% of the distance. The strongest ...


3

If the cable of the "elevator" is not connected to a point on earth, then the satellite must be in a geostationary orbit (or it will float away); this implies that if you now attach something to the platform (increasing the pull on the cable) you will pull the satellite down to earth. And as @lionelbrits pointed out, the pulling part of a space elevator ...



Top 50 recent answers are included