You land on an earth size planet where the gravity is the same as earth so the density is the same as earth.
Now the shuttle can take off because it pulls about 2and1/2 G's .
But something horrible is happening. The planet for some unknown reason is shrinking, maybe eventually it will be a black hole and there is no escape. They have engineered the shuttle to get 5 G's of accel but for half the time. What is the smallest the planet can shrink, the density increasing all the time, and they still get to take off? We can call the planet 13,000 Km in diameter to start, about the same as earth.
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Originally posted by sonhousetake it to the Science forum spanky...I always wanted to say that...just kidding😉
You land on an earth size planet where the gravity is the same as earth so the density is the same as earth.
Now the shuttle can take off because it pulls about 2and1/2 G's .
But something horrible is happening. The planet for some unknown reason is shrinking, maybe eventually it will be a black hole and there is no escape. They have engineered the shuttl ...[text shortened]... t to take off? We can call the planet 13,000 Km in diameter to start, about the same as earth.
Originally posted by sonhouseGravity will not change in the slightest for the shuttle. The only difference is it will start from closer to the center if it waits to take off.
You land on an earth size planet where the gravity is the same as earth so the density is the same as earth.
Now the shuttle can take off because it pulls about 2and1/2 G's .
But something horrible is happening. The planet for some unknown reason is shrinking, maybe eventually it will be a black hole and there is no escape. They have engineered the shuttl ...[text shortened]... t to take off? We can call the planet 13,000 Km in diameter to start, about the same as earth.
Originally posted by AThousandYoungThe gravitational force depends on the distance between the centre-of-masses.
Gravity will not change in the slightest for the shuttle. The only difference is it will start from closer to the center if it waits to take off.
That would mean gravitational force would increase as the planet is shrinking.
Gravity will not change in the slightest for the shuttle. The only difference is it will start from closer to the center if it waits to take off.
Well, if it takes off immediately, all it needs is >1 G. Then, as I think you meant to say, gravitational force will not increase because neither mass changes, nor does the distance between them ever decrease. So then it makes no difference how small the planet becomes (even if it becomes a black hole). This is, of course, assuming that the planet/rocket system is isolated.
Then again, if it does wait to take off, then we still need to know those two pieces of information mentioned earlier, as well as exactly how long the rocket waits to take off.
Originally posted by JirakonThey do have a problem with time, it is shrinking at an alarming rate, in about one hour it will require 10 G's so they better get busy, they have about 30 minutes if they want to get off with 5 G's. The Shuttle main engines are good for about 15 minutes of thrust.
Gravity will not change in the slightest for the shuttle. The only difference is it will start from closer to the center if it waits to take off.
Well, if it takes off immediately, all it needs is >1 G. Then, as I think you meant to say, gravitational force will not increase because neither mass changes, nor does the distance between them ever decr ...[text shortened]... eces of information mentioned earlier, as well as exactly how long the rocket waits to take off.
I agree with the others.
Providing it is accelerating away at greater than 1G it will escape. The shrinking planet is not getting more massive therefore its pull on the shuttle is the same.
(Even if it shrinks to a black hole the gravitational pull at a distance equivalent to the radius of the original planet will still be 1G)
A trickier problem is taking off from a planet which is growing in mass ......
Wait a little here...
If you increase the density, thus shrinking it, of the Earth, until you get a diameter in the size of 8 millimeter, then you'll have a Black Hole. If you are at the edge of a black hole, even with this tiny size, then you will not be able to lift off anymore.
Where, in this reasoning, am I wrong?
Originally posted by FabianFnasYour thinking is fine. If the mass of a planet doesn't change, and it starts shrinking for some unknown or unknowable reason, maybe an alien race is zapping the planet with a cosmic heatshrink gun, whatever, the situation is the planet is getting smaller but the mass does not change. The density goes up, so what mass there is gets concentrated into a smaller volume, so gravity is a volumetric beast, the surface gravity will go up. It HAS to go up.
Wait a little here...
If you increase the density, thus shrinking it, of the Earth, until you get a diameter in the size of 8 millimeter, then you'll have a Black Hole. If you are at the edge of a black hole, even with this tiny size, then you will not be able to lift off anymore.
Where, in this reasoning, am I wrong?
Originally posted by sonhouseYou know of Gauss' law? If you change only the density of the planet the gravitational field won't be affected at all at the same point.
You land on an earth size planet where the gravity is the same as earth so the density is the same as earth.
Now the shuttle can take off because it pulls about 2and1/2 G's .
But something horrible is happening. The planet for some unknown reason is shrinking, maybe eventually it will be a black hole and there is no escape. They have engineered the shuttl t to take off? We can call the planet 13,000 Km in diameter to start, about the same as earth.
The gravitational field on the surface of the planet will indeed increase (proportional to the density^2/3 ), but only because the distance to the center of mass of the planet decreases like density^1/3.
Gauss' law argument: draw a sphere which encapsulates the whole surface of the original planet. Find the flux. If you diminish the size of the planet, maintaining spherical symmetry, the flux will be the same everywhere in that sphere's surface, therefore the field will be the same too.
Actually you don't have to maintain spherical symmetry for the argument to be true... But it'a a lot easier to visualize keeping the symmetry. No integration over strange surfaces needed 🙂
EDIT: this was done with classical physics. If you want general relativistic effects, the whole story is a lot different. It's a good exercise 🙂 Gonna get back to it later with general relativistic calculations.
Originally posted by FabianFnasI was considering the situation where the spaceship takes off at the moment (or before) the planet starts shrinking. As Seigado points out the efect on the spaceship is unchanged.
Wait a little here...
If you increase the density, thus shrinking it, of the Earth, until you get a diameter in the size of 8 millimeter, then you'll have a Black Hole. If you are at the edge of a black hole, even with this tiny size, then you will not be able to lift off anymore.
Where, in this reasoning, am I wrong?
IF HOWEVER the spaceship delays take-off then it gets closer to the centre of the planet so obviously the gravity increases.
TWO situations - lets not confuse them.
Originally posted by wolfgang59I made my point before Seigado made his. I hadn't had a chance reading his. I wasn't referring ho him (unless I have a time travelling device at hand).
I was considering the situation where the spaceship takes off at the moment (or before) the planet starts shrinking. As Seigado points out the efect on the spaceship is unchanged.
IF HOWEVER the spaceship delays take-off then it gets closer to the centre of the planet so obviously the gravity increases.
TWO situations - lets not confuse them.
I assume that the spaceship doesn't hoover above the surface at the same distance from center before it takes off? I assume simply that it follows the surface downwards toward the center. If not, there is not a problem, is it?
Originally posted by FabianFnas😕
I made my point before Seigado made his. I hadn't had a chance reading his. I wasn't referring ho him (unless I have a time travelling device at hand).
I assume that the spaceship doesn't hoover above the surface at the same distance from center before it takes off? I assume simply that it follows the surface downwards toward the center. If not, there is not a problem, is it?
I was just justifying my position and used Serigados excellent explanayion to help.
I never meant to suggest that you had read his post did I?
Originally posted by wolfgang59If you'd read the future postings in this thread I'm sure that you would answer differently... 😵
😕
I was just justifying my position and used Serigados excellent explanayion to help.
I never meant to suggest that you had read his post did I?