What's the largest planet you can jump off and leave orbit?

Answer 1

Nobody has given you a real answer yet so here you go:

The smallest planet you could jump away from and escape its gravitational field would be roughly 1.5km in radius and have a mass of about 55,000,000,000,000kg (about one hundred billionth the mass of the earth)

Here is how I calculated it:
"In physics, for a given gravitational field and a given position, the escape velocity is the minimum speed an object without propulsion, at that position, needs to have to move away indefinitely from the source of the field, as opposed to falling back or staying in an orbit within a bounded distance from the source."(http://en.wikipedia.org/wiki/Escape_velocity)

When you say jump and escape the planet, the escape velocity is the minimum jump speed you would need for this, so let's assume you're very good at jumping and can jump a meter on earth, then get your approximate jumping velocity.

vf²=vi²+2ad
vf=final velocity, vi=initial velocity, a=acceleration, d=distance
(all in S.I. units)

Set this up so we are looking at the top of your jump where you stop, so vf=0, d=1m

plug in and solve

0=vi+2(-9.8)1
vi=4.5m/s (about)

so you can jump at about 4.5m/s.. roughly

Now let's move to the formula for escape velocity.

Ve=√(2GM/r)

where ve is the escape velocity, G is the gravitational constant, M is the mass of the body being escaped from, and r is the distance between the centre of the body and the point at which escape velocity is being calculated.

Now we have a little bit of a problem, we don't know the size of your planet, but we can figure that out by deciding what it's going to be made of. We can guess it's made of medium density rock and has a mass M of 3,000kg/m³. So with V as volume in cubic metres, M=V(3,000kg).

Now we can relate r and M. Assuming your planet is spherical, we can say V=4/3πr³, and M=(3000kg)4/3πr³.

Substitute the M into the ve formula to get

Ve=√(2G[(3000kg)4/3πr³]/r)

simplify

Ve=√(8/3πG(3000kg)r²)

now substitute the 4.5m/s jumping velocity in and solve for r

4.5=√(8/3πG(3000kg)r²)
4.5=8/3πG(3000kg)r²
4.5=8/3πG(3000kg)r²
(4.5)x3/(8πG(3000kg))=r²
r=1640m

now let's get the mass M from r
M=(3000kg)4/3πr³
M= 55,000,000,000,000kg (about one hundred billionth the mass of the earth)

This was based on your jumping velocity being about 4.5m/s, no air resistance, the planet's density being 3,000kg/m³. Escape velocity is independent of the direction of the velocity(except of course if you jumped tangentially to the surface and hit a mountain or something silly like that) so jumping straight up does not matter.

Answer 2

I like Michael A's answer, but would expand on it a bit:
if you jump high enough, you might get into the orbit of ANOTHER planet! For example, if you were on the moon (which as Michael points out) could be a planet, and jumped really high, you could get to a point where earth is pulling on you harder than the moon is.
But really Michael is still right because to some extent the earth is orbiting the moon, and the moon is orbiting the earth, and we're all orbiting the sun...

Answer 3

WEll, by the strict definition of a planet (a body whose gravity pulls it into a sphere), none.

But lets consider smaller bodies - planetoids if you like.

Lets assume that you could put the same energy into a jump as on Earth, and that the density of the planetoid is the same as Earths. Then we can just equate the potential energy of the person on the planetoid to the energy they can put into a jump.

From this you can show the radius of the planetoid would be the square root of Re^2 * (h / [Re + h)]) - Re being the radius of the Earth and h the height you can jump on Earth.

The best high jumper can jump about 2.4m.

Doing the numbers the planetoid would have a radius of about 4 km.

Answer 4

Well I think I understand his question.
We can use physics to solve this problem! Too all those who are looking at this. Yeah I have lots of free time!
**********Part 1
According to Wikipedia:
Highest jump: Javier Sotomayor, 8ft 1/2in (2.45 m)

What we first want to know is the maximum velocity an individual can jump at. We can use physics motion laws to calculate the intial velocity. Of course we have to make a few more assumptions. The law that we will use is as follows:

----Assumption:
1.) No wind resistance
2.) He jumps straight going up
----Assumption:

Equation:
(Vf)²=(Vi)² + 2*a*s
Where:
(Vf)=Final Velocity (0; at max height our velocity is zero)
(Vi)=Initial Velcoity (What we are trying to find)
a = Acceleration Due to gravity (Earth's acceleration =9.81m/s²)
s = Distance travelled (2.45m)

So...
(Vf)²=(Vi)² + 2*a*s
0 =(Vi)² + 2*-9.81m/s² * 2.45m
0 =(Vi)² - 48.069
48.069 = (Vi)²
Vi = 6.9332m/s!!!
Note:Use a negative sign in front of gravity b/c when Javier jumps he's really decelerating

**********Part II
Now for part 2 we have to use escape velocity equations:
Ve=sqrt{2g*r}.

Where
Ve = escape velocity
g = gravity of planet
r = radius of planet

6.9332² = 2*g*r
48.069 = 2*g*r
24.0345 >= g*r
So the escape velocity depends on the gravity of the planet and it's radius! If 24.0345 is greater than or equal to the gravity of the planet times the radius of the planet, then you( I mean Javier) will fly off into space...


Find the gravity and the radius of a planet and match it to see if you escape!

Answer 5

Everything has its force of Gravity (Newton's Law Of Gravitation).
A small rock is also like a planet. But, we don't consider it a planet. If you have studied Classical Mechanics, you would know the fact that everything is in freefall with everything or simply everything is attracting everything. So, if you are 100 miles away from the rock, it is still attracting you. Though it's strength of gravity is very weak, it is attracting you. So, your question makes no sense.

Answer 6

I think you'll find that 'any' is the right answer, because of the definition of a planet being a huge piece of rock

Answer 7

you can't jump off any planet and leave orbit.The larger the planet the stronger the gravitational pull, but you still can't ''jump'' off any planet and leave orbit

Answer 8

Would that be planetary orbit, or solar orbit? I can do the first already. Am working on holding my breath.

Answer 9

Uranus.

Answer 10

Personally none. Theoretically any.