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Largest radius sphere with Earth's surface gravity on which you could jump at escape velocity? Bigger than B612?
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In the book Le Petit Prince by "French aristocrat, writer, poet, and pioneering aviator Antoine de Saint-Exupéry" the main character lives on an extremely tiny asteroid with the name B612. It is this "asteroid" that the B612 Foundation was named after, founded by retired astronaut and entrepreneur Dr. Ed Lu and Drs. Clark Chapman and Piet Hut.
What "escapes" this XKCD-based answer (pardon the pun) is that it is the $fracmassradius$ ratio that is key to the escape velocity, not just the surface gravity.
$$v_esc = sqrtleft(frac2 GMr_0 right)$$
Question: Using numbers from my answer there, what would be the largest radius sphere (and corresponding mass) that had Earth's surface gravity of about 9.8 m/s^2 that you could jump at escape velocity?
"Bonus points" for an approximate scale height of an atmopshere on the unusual theoretical body in your answer (not B612 of course)
Source
physics escape-velocity
asked 4 hours ago
uhoh
30.8k15106380
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up vote
3
down vote
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In the book Le Petit Prince by "French aristocrat, writer, poet, and pioneering aviator Antoine de Saint-Exupéry" the main character lives on an extremely tiny asteroid with the name B612. It is this "asteroid" that the B612 Foundation was named after, founded by retired astronaut and entrepreneur Dr. Ed Lu and Drs. Clark Chapman and Piet Hut.
What "escapes" this XKCD-based answer (pardon the pun) is that it is the $fracmassradius$ ratio that is key to the escape velocity, not just the surface gravity.
$$v_esc = sqrtleft(frac2 GMr_0 right)$$
Question: Using numbers from my answer there, what would be the largest radius sphere (and corresponding mass) that had Earth's surface gravity of about 9.8 m/s^2 that you could jump at escape velocity?
"Bonus points" for an approximate scale height of an atmopshere on the unusual theoretical body in your answer (not B612 of course)
Source
physics escape-velocity
asked 4 hours ago
uhoh
30.8k15106380
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
In the book Le Petit Prince by "French aristocrat, writer, poet, and pioneering aviator Antoine de Saint-Exupéry" the main character lives on an extremely tiny asteroid with the name B612. It is this "asteroid" that the B612 Foundation was named after, founded by retired astronaut and entrepreneur Dr. Ed Lu and Drs. Clark Chapman and Piet Hut.
What "escapes" this XKCD-based answer (pardon the pun) is that it is the $fracmassradius$ ratio that is key to the escape velocity, not just the surface gravity.
$$v_esc = sqrtleft(frac2 GMr_0 right)$$
Question: Using numbers from my answer there, what would be the largest radius sphere (and corresponding mass) that had Earth's surface gravity of about 9.8 m/s^2 that you could jump at escape velocity?
"Bonus points" for an approximate scale height of an atmopshere on the unusual theoretical body in your answer (not B612 of course)
Source
physics escape-velocity
asked 4 hours ago
uhoh
30.8k15106380
In the book Le Petit Prince by "French aristocrat, writer, poet, and pioneering aviator Antoine de Saint-Exupéry" the main character lives on an extremely tiny asteroid with the name B612. It is this "asteroid" that the B612 Foundation was named after, founded by retired astronaut and entrepreneur Dr. Ed Lu and Drs. Clark Chapman and Piet Hut.
What "escapes" this XKCD-based answer (pardon the pun) is that it is the $fracmassradius$ ratio that is key to the escape velocity, not just the surface gravity.
$$v_esc = sqrtleft(frac2 GMr_0 right)$$
Question: Using numbers from my answer there, what would be the largest radius sphere (and corresponding mass) that had Earth's surface gravity of about 9.8 m/s^2 that you could jump at escape velocity?
"Bonus points" for an approximate scale height of an atmopshere on the unusual theoretical body in your answer (not B612 of course)
Source
physics escape-velocity
physics escape-velocity
asked 4 hours ago
uhoh
30.8k15106380
asked 4 hours ago
uhoh
30.8k15106380
edited 3 hours ago
asked 4 hours ago
uhoh
30.8k15106380
asked 4 hours ago
uhoh
30.8k15106380
asked 4 hours ago
uhoh
30.8k15106380
30.8k15106380
add a comment |Â
add a comment |Â
1 Answer
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This is quite a nice calculation. Suppose someone can jump (more exactly raise their centre of mass) $h$ meters on Earth. Then $$mgh = 1/2 m v^2$$. So, if $v > v_esc$ then
$$mgh > fracm2 frac2GMr$$
so $$gh > fracGMr$$
However we know that our planetoid has surface gravity $g$ so
$$g = fracGMr^2$$
Combining these, $$fracGMhr^2 > fracGMr$$ or
$$h > r$$
An Olympic high jumper probably lifts their centre of mass about $1m$ so $r$ is certainly going to be of this order. Now you have to worry about your definitions. Is your centre of mass on the surface, or your feet at the start, for instance, but a small number of meters is the general range.
Taking $r = 1$ we get $9.8 = GM$ so $M = 1.4 times 10^11$ and the density is about $10^10 kg/m^3$ a bit denser than white dwarf material, but nowhere near as dense as neutronium.
answered 2 hours ago
Steve Linton
4,8411632
oh, this is a really nice answer!
– uhoh
1 hour ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
This is quite a nice calculation. Suppose someone can jump (more exactly raise their centre of mass) $h$ meters on Earth. Then $$mgh = 1/2 m v^2$$. So, if $v > v_esc$ then
$$mgh > fracm2 frac2GMr$$
so $$gh > fracGMr$$
However we know that our planetoid has surface gravity $g$ so
$$g = fracGMr^2$$
Combining these, $$fracGMhr^2 > fracGMr$$ or
$$h > r$$
An Olympic high jumper probably lifts their centre of mass about $1m$ so $r$ is certainly going to be of this order. Now you have to worry about your definitions. Is your centre of mass on the surface, or your feet at the start, for instance, but a small number of meters is the general range.
Taking $r = 1$ we get $9.8 = GM$ so $M = 1.4 times 10^11$ and the density is about $10^10 kg/m^3$ a bit denser than white dwarf material, but nowhere near as dense as neutronium.
answered 2 hours ago
Steve Linton
4,8411632
oh, this is a really nice answer!
– uhoh
1 hour ago
add a comment |Â
up vote
4
down vote
This is quite a nice calculation. Suppose someone can jump (more exactly raise their centre of mass) $h$ meters on Earth. Then $$mgh = 1/2 m v^2$$. So, if $v > v_esc$ then
$$mgh > fracm2 frac2GMr$$
so $$gh > fracGMr$$
However we know that our planetoid has surface gravity $g$ so
$$g = fracGMr^2$$
Combining these, $$fracGMhr^2 > fracGMr$$ or
$$h > r$$
An Olympic high jumper probably lifts their centre of mass about $1m$ so $r$ is certainly going to be of this order. Now you have to worry about your definitions. Is your centre of mass on the surface, or your feet at the start, for instance, but a small number of meters is the general range.
Taking $r = 1$ we get $9.8 = GM$ so $M = 1.4 times 10^11$ and the density is about $10^10 kg/m^3$ a bit denser than white dwarf material, but nowhere near as dense as neutronium.
answered 2 hours ago
Steve Linton
4,8411632
oh, this is a really nice answer!
– uhoh
1 hour ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
This is quite a nice calculation. Suppose someone can jump (more exactly raise their centre of mass) $h$ meters on Earth. Then $$mgh = 1/2 m v^2$$. So, if $v > v_esc$ then
$$mgh > fracm2 frac2GMr$$
so $$gh > fracGMr$$
However we know that our planetoid has surface gravity $g$ so
$$g = fracGMr^2$$
Combining these, $$fracGMhr^2 > fracGMr$$ or
$$h > r$$
An Olympic high jumper probably lifts their centre of mass about $1m$ so $r$ is certainly going to be of this order. Now you have to worry about your definitions. Is your centre of mass on the surface, or your feet at the start, for instance, but a small number of meters is the general range.
Taking $r = 1$ we get $9.8 = GM$ so $M = 1.4 times 10^11$ and the density is about $10^10 kg/m^3$ a bit denser than white dwarf material, but nowhere near as dense as neutronium.
answered 2 hours ago
Steve Linton
4,8411632
This is quite a nice calculation. Suppose someone can jump (more exactly raise their centre of mass) $h$ meters on Earth. Then $$mgh = 1/2 m v^2$$. So, if $v > v_esc$ then
$$mgh > fracm2 frac2GMr$$
so $$gh > fracGMr$$
However we know that our planetoid has surface gravity $g$ so
$$g = fracGMr^2$$
Combining these, $$fracGMhr^2 > fracGMr$$ or
$$h > r$$
An Olympic high jumper probably lifts their centre of mass about $1m$ so $r$ is certainly going to be of this order. Now you have to worry about your definitions. Is your centre of mass on the surface, or your feet at the start, for instance, but a small number of meters is the general range.
Taking $r = 1$ we get $9.8 = GM$ so $M = 1.4 times 10^11$ and the density is about $10^10 kg/m^3$ a bit denser than white dwarf material, but nowhere near as dense as neutronium.
answered 2 hours ago
Steve Linton
4,8411632
answered 2 hours ago
Steve Linton
4,8411632
answered 2 hours ago
Steve Linton
4,8411632
answered 2 hours ago
Steve Linton
4,8411632
4,8411632
oh, this is a really nice answer!
– uhoh
1 hour ago
add a comment |Â
oh, this is a really nice answer!
– uhoh
1 hour ago
oh, this is a really nice answer!
– uhoh
1 hour ago
oh, this is a really nice answer!
– uhoh
1 hour ago
add a comment |Â
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