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Could a human jump off Mimas without return?
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A similar question has already been asked on dwarf planet Ceres:
Could a Human reach escape velocity by jumping from the surface of Ceres (a dwarf planet)?
Ceres has 2.9% of Earth's gravity. Saturn's moon Mimas has 0.6% of Earth's gravity. If you jump strong enough, could you fly off into outer space from Mimas?
launch gravity physics escape-velocity saturn
asked 3 hours ago
user27822
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up vote
3
down vote
favorite
A similar question has already been asked on dwarf planet Ceres:
Could a Human reach escape velocity by jumping from the surface of Ceres (a dwarf planet)?
Ceres has 2.9% of Earth's gravity. Saturn's moon Mimas has 0.6% of Earth's gravity. If you jump strong enough, could you fly off into outer space from Mimas?
launch gravity physics escape-velocity saturn
asked 3 hours ago
user27822
1839
New contributor
user27822 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
A similar question has already been asked on dwarf planet Ceres:
Could a Human reach escape velocity by jumping from the surface of Ceres (a dwarf planet)?
Ceres has 2.9% of Earth's gravity. Saturn's moon Mimas has 0.6% of Earth's gravity. If you jump strong enough, could you fly off into outer space from Mimas?
launch gravity physics escape-velocity saturn
asked 3 hours ago
user27822
1839
New contributor
user27822 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
A similar question has already been asked on dwarf planet Ceres:
Could a Human reach escape velocity by jumping from the surface of Ceres (a dwarf planet)?
Ceres has 2.9% of Earth's gravity. Saturn's moon Mimas has 0.6% of Earth's gravity. If you jump strong enough, could you fly off into outer space from Mimas?
launch gravity physics escape-velocity saturn
launch gravity physics escape-velocity saturn
asked 3 hours ago
user27822
1839
New contributor
user27822 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
user27822
1839
New contributor
user27822 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
user27822
1839
New contributor
user27822 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
user27822
1839
asked 3 hours ago
user27822
1839
1839
New contributor
user27822 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user27822 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user27822 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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2 Answers
2
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oldest
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up vote
4
down vote
No. Mimas' gravity is 0.064 m/s2, you need gravity to be about 1/20 of that to escape using a bike and a ramp (going by Deimos' surface gravity of 1/20 that of Mimas), lower still to escape by jumping :
answered 2 hours ago
Hobbes
80.2k2220360
While more precisely, Mimas has 0.0636 m/s^2 or about 0.209 ft/s^2 which is about 0.006 g.
– user27822
1 hour ago
Too late, I can't delete an accepted answer.
– Hobbes
35 mins ago
add a comment |Â
up vote
0
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tl;dr: No chance, not even close!
The escape velocity from the surface of a round (spherically symmetric) body is given by
$$v_esc = sqrtleft(frac2 GMr_0 right), $$
showing that it is the $fracmassradius$ ratio that's key here, not just the surface gravity given by
$$a_g = -fracGMr_0^2. $$
So since
$$v_esc = sqrta_g r_0, $$
a lower density but larger radius body with the same surface gravity would have a higher escape velocity. You can think of that as "the gravity extending outward farther" or better yet, just dropping off slower. Gravity drops by a factor of 4 at $2r_0$, so if $r_0$ is bigger, so is $2r_0$.
The problem is a little tougher because you have to look at the design of human legs. They are optimized to work in Earth gravity; they have mass an moments of inertia that work with muscle strength and the speed with which muscle fibers can contract. For that you can start with this excellent answer to bibliography for the question Any scholarly or serious work in Sports Science for the low surface gravity of Mars or the Moon? or other things tagged reduced-gravity-sports.
Let's look at what happens on Earth. Most people will find it a challenge to get to 1 meter in a standing high jump, and the world's record is 1.65 meters. Let's use 70 kg and 1 meter at $g_0$=9.8 m/s^2, some basic kinematics, and this page linking to the PDF Optimum Take-Off Range in Vertical Jumping to get a better picture
Source
Source
Published article Analysis of standing vertical jumps using a force platform Nicholas Linthorne, UNSW.
There's about a 1000 Newton force beyond the ~750 N supporting weight against gravity, or about 14 m/s^2, for about 0.25 meters. That's about 0.19 seconds and a take-off velocity of 2.6 m/s using $v = sqrt2 g h$ and $t = sqrt2x/g$.
The surface gravities of Mimas and Ceres are 0.064 and 0.28 m/s^2 respectively, and their escape velocities are
If you could develop only the 1000 Newtons over 0.25 meters at those surface gravities, you would also achieve that ~2.6 m/s velocity.
However, their escape velocities are 160 and 510 m/s, respectively! So... no chance, not even close!
answered 1 hour ago
uhoh
30.7k15106380
I've asked the follow-up question Largest radius sphere with Earth's surface gravity on which you could jump at escape velocity? Bigger than B612?
– uhoh
19 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
No. Mimas' gravity is 0.064 m/s2, you need gravity to be about 1/20 of that to escape using a bike and a ramp (going by Deimos' surface gravity of 1/20 that of Mimas), lower still to escape by jumping :
answered 2 hours ago
Hobbes
80.2k2220360
While more precisely, Mimas has 0.0636 m/s^2 or about 0.209 ft/s^2 which is about 0.006 g.
– user27822
1 hour ago
Too late, I can't delete an accepted answer.
– Hobbes
35 mins ago
add a comment |Â
up vote
4
down vote
No. Mimas' gravity is 0.064 m/s2, you need gravity to be about 1/20 of that to escape using a bike and a ramp (going by Deimos' surface gravity of 1/20 that of Mimas), lower still to escape by jumping :
answered 2 hours ago
Hobbes
80.2k2220360
While more precisely, Mimas has 0.0636 m/s^2 or about 0.209 ft/s^2 which is about 0.006 g.
– user27822
1 hour ago
Too late, I can't delete an accepted answer.
– Hobbes
35 mins ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
No. Mimas' gravity is 0.064 m/s2, you need gravity to be about 1/20 of that to escape using a bike and a ramp (going by Deimos' surface gravity of 1/20 that of Mimas), lower still to escape by jumping :
answered 2 hours ago
Hobbes
80.2k2220360
No. Mimas' gravity is 0.064 m/s2, you need gravity to be about 1/20 of that to escape using a bike and a ramp (going by Deimos' surface gravity of 1/20 that of Mimas), lower still to escape by jumping :
answered 2 hours ago
Hobbes
80.2k2220360
edited 25 mins ago
answered 2 hours ago
Hobbes
80.2k2220360
answered 2 hours ago
Hobbes
80.2k2220360
answered 2 hours ago
Hobbes
80.2k2220360
80.2k2220360
While more precisely, Mimas has 0.0636 m/s^2 or about 0.209 ft/s^2 which is about 0.006 g.
– user27822
1 hour ago
Too late, I can't delete an accepted answer.
– Hobbes
35 mins ago
add a comment |Â
While more precisely, Mimas has 0.0636 m/s^2 or about 0.209 ft/s^2 which is about 0.006 g.
– user27822
1 hour ago
Too late, I can't delete an accepted answer.
– Hobbes
35 mins ago
While more precisely, Mimas has 0.0636 m/s^2 or about 0.209 ft/s^2 which is about 0.006 g.
– user27822
1 hour ago
While more precisely, Mimas has 0.0636 m/s^2 or about 0.209 ft/s^2 which is about 0.006 g.
– user27822
1 hour ago
Too late, I can't delete an accepted answer.
– Hobbes
35 mins ago
Too late, I can't delete an accepted answer.
– Hobbes
35 mins ago
add a comment |Â
up vote
0
down vote
tl;dr: No chance, not even close!
The escape velocity from the surface of a round (spherically symmetric) body is given by
$$v_esc = sqrtleft(frac2 GMr_0 right), $$
showing that it is the $fracmassradius$ ratio that's key here, not just the surface gravity given by
$$a_g = -fracGMr_0^2. $$
So since
$$v_esc = sqrta_g r_0, $$
a lower density but larger radius body with the same surface gravity would have a higher escape velocity. You can think of that as "the gravity extending outward farther" or better yet, just dropping off slower. Gravity drops by a factor of 4 at $2r_0$, so if $r_0$ is bigger, so is $2r_0$.
The problem is a little tougher because you have to look at the design of human legs. They are optimized to work in Earth gravity; they have mass an moments of inertia that work with muscle strength and the speed with which muscle fibers can contract. For that you can start with this excellent answer to bibliography for the question Any scholarly or serious work in Sports Science for the low surface gravity of Mars or the Moon? or other things tagged reduced-gravity-sports.
Let's look at what happens on Earth. Most people will find it a challenge to get to 1 meter in a standing high jump, and the world's record is 1.65 meters. Let's use 70 kg and 1 meter at $g_0$=9.8 m/s^2, some basic kinematics, and this page linking to the PDF Optimum Take-Off Range in Vertical Jumping to get a better picture
Source
Source
Published article Analysis of standing vertical jumps using a force platform Nicholas Linthorne, UNSW.
There's about a 1000 Newton force beyond the ~750 N supporting weight against gravity, or about 14 m/s^2, for about 0.25 meters. That's about 0.19 seconds and a take-off velocity of 2.6 m/s using $v = sqrt2 g h$ and $t = sqrt2x/g$.
The surface gravities of Mimas and Ceres are 0.064 and 0.28 m/s^2 respectively, and their escape velocities are
If you could develop only the 1000 Newtons over 0.25 meters at those surface gravities, you would also achieve that ~2.6 m/s velocity.
However, their escape velocities are 160 and 510 m/s, respectively! So... no chance, not even close!
answered 1 hour ago
uhoh
30.7k15106380
I've asked the follow-up question Largest radius sphere with Earth's surface gravity on which you could jump at escape velocity? Bigger than B612?
– uhoh
19 mins ago
add a comment |Â
up vote
0
down vote
tl;dr: No chance, not even close!
The escape velocity from the surface of a round (spherically symmetric) body is given by
$$v_esc = sqrtleft(frac2 GMr_0 right), $$
showing that it is the $fracmassradius$ ratio that's key here, not just the surface gravity given by
$$a_g = -fracGMr_0^2. $$
So since
$$v_esc = sqrta_g r_0, $$
a lower density but larger radius body with the same surface gravity would have a higher escape velocity. You can think of that as "the gravity extending outward farther" or better yet, just dropping off slower. Gravity drops by a factor of 4 at $2r_0$, so if $r_0$ is bigger, so is $2r_0$.
The problem is a little tougher because you have to look at the design of human legs. They are optimized to work in Earth gravity; they have mass an moments of inertia that work with muscle strength and the speed with which muscle fibers can contract. For that you can start with this excellent answer to bibliography for the question Any scholarly or serious work in Sports Science for the low surface gravity of Mars or the Moon? or other things tagged reduced-gravity-sports.
Let's look at what happens on Earth. Most people will find it a challenge to get to 1 meter in a standing high jump, and the world's record is 1.65 meters. Let's use 70 kg and 1 meter at $g_0$=9.8 m/s^2, some basic kinematics, and this page linking to the PDF Optimum Take-Off Range in Vertical Jumping to get a better picture
Source
Source
Published article Analysis of standing vertical jumps using a force platform Nicholas Linthorne, UNSW.
There's about a 1000 Newton force beyond the ~750 N supporting weight against gravity, or about 14 m/s^2, for about 0.25 meters. That's about 0.19 seconds and a take-off velocity of 2.6 m/s using $v = sqrt2 g h$ and $t = sqrt2x/g$.
The surface gravities of Mimas and Ceres are 0.064 and 0.28 m/s^2 respectively, and their escape velocities are
If you could develop only the 1000 Newtons over 0.25 meters at those surface gravities, you would also achieve that ~2.6 m/s velocity.
However, their escape velocities are 160 and 510 m/s, respectively! So... no chance, not even close!
answered 1 hour ago
uhoh
30.7k15106380
I've asked the follow-up question Largest radius sphere with Earth's surface gravity on which you could jump at escape velocity? Bigger than B612?
– uhoh
19 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
tl;dr: No chance, not even close!
The escape velocity from the surface of a round (spherically symmetric) body is given by
$$v_esc = sqrtleft(frac2 GMr_0 right), $$
showing that it is the $fracmassradius$ ratio that's key here, not just the surface gravity given by
$$a_g = -fracGMr_0^2. $$
So since
$$v_esc = sqrta_g r_0, $$
a lower density but larger radius body with the same surface gravity would have a higher escape velocity. You can think of that as "the gravity extending outward farther" or better yet, just dropping off slower. Gravity drops by a factor of 4 at $2r_0$, so if $r_0$ is bigger, so is $2r_0$.
The problem is a little tougher because you have to look at the design of human legs. They are optimized to work in Earth gravity; they have mass an moments of inertia that work with muscle strength and the speed with which muscle fibers can contract. For that you can start with this excellent answer to bibliography for the question Any scholarly or serious work in Sports Science for the low surface gravity of Mars or the Moon? or other things tagged reduced-gravity-sports.
Let's look at what happens on Earth. Most people will find it a challenge to get to 1 meter in a standing high jump, and the world's record is 1.65 meters. Let's use 70 kg and 1 meter at $g_0$=9.8 m/s^2, some basic kinematics, and this page linking to the PDF Optimum Take-Off Range in Vertical Jumping to get a better picture
Source
Source
Published article Analysis of standing vertical jumps using a force platform Nicholas Linthorne, UNSW.
There's about a 1000 Newton force beyond the ~750 N supporting weight against gravity, or about 14 m/s^2, for about 0.25 meters. That's about 0.19 seconds and a take-off velocity of 2.6 m/s using $v = sqrt2 g h$ and $t = sqrt2x/g$.
The surface gravities of Mimas and Ceres are 0.064 and 0.28 m/s^2 respectively, and their escape velocities are
If you could develop only the 1000 Newtons over 0.25 meters at those surface gravities, you would also achieve that ~2.6 m/s velocity.
However, their escape velocities are 160 and 510 m/s, respectively! So... no chance, not even close!
answered 1 hour ago
uhoh
30.7k15106380
tl;dr: No chance, not even close!
The escape velocity from the surface of a round (spherically symmetric) body is given by
$$v_esc = sqrtleft(frac2 GMr_0 right), $$
showing that it is the $fracmassradius$ ratio that's key here, not just the surface gravity given by
$$a_g = -fracGMr_0^2. $$
So since
$$v_esc = sqrta_g r_0, $$
a lower density but larger radius body with the same surface gravity would have a higher escape velocity. You can think of that as "the gravity extending outward farther" or better yet, just dropping off slower. Gravity drops by a factor of 4 at $2r_0$, so if $r_0$ is bigger, so is $2r_0$.
The problem is a little tougher because you have to look at the design of human legs. They are optimized to work in Earth gravity; they have mass an moments of inertia that work with muscle strength and the speed with which muscle fibers can contract. For that you can start with this excellent answer to bibliography for the question Any scholarly or serious work in Sports Science for the low surface gravity of Mars or the Moon? or other things tagged reduced-gravity-sports.
Let's look at what happens on Earth. Most people will find it a challenge to get to 1 meter in a standing high jump, and the world's record is 1.65 meters. Let's use 70 kg and 1 meter at $g_0$=9.8 m/s^2, some basic kinematics, and this page linking to the PDF Optimum Take-Off Range in Vertical Jumping to get a better picture
Source
Source
Published article Analysis of standing vertical jumps using a force platform Nicholas Linthorne, UNSW.
There's about a 1000 Newton force beyond the ~750 N supporting weight against gravity, or about 14 m/s^2, for about 0.25 meters. That's about 0.19 seconds and a take-off velocity of 2.6 m/s using $v = sqrt2 g h$ and $t = sqrt2x/g$.
The surface gravities of Mimas and Ceres are 0.064 and 0.28 m/s^2 respectively, and their escape velocities are
If you could develop only the 1000 Newtons over 0.25 meters at those surface gravities, you would also achieve that ~2.6 m/s velocity.
However, their escape velocities are 160 and 510 m/s, respectively! So... no chance, not even close!
answered 1 hour ago
uhoh
30.7k15106380
edited 1 hour ago
answered 1 hour ago
uhoh
30.7k15106380
answered 1 hour ago
uhoh
30.7k15106380
answered 1 hour ago
uhoh
30.7k15106380
30.7k15106380
I've asked the follow-up question Largest radius sphere with Earth's surface gravity on which you could jump at escape velocity? Bigger than B612?
– uhoh
19 mins ago
add a comment |Â
I've asked the follow-up question Largest radius sphere with Earth's surface gravity on which you could jump at escape velocity? Bigger than B612?
– uhoh
19 mins ago
I've asked the follow-up question Largest radius sphere with Earth's surface gravity on which you could jump at escape velocity? Bigger than B612?
– uhoh
19 mins ago
I've asked the follow-up question Largest radius sphere with Earth's surface gravity on which you could jump at escape velocity? Bigger than B612?
– uhoh
19 mins ago
add a comment |Â
user27822 is a new contributor. Be nice, and check out our Code of Conduct.
user27822 is a new contributor. Be nice, and check out our Code of Conduct.
user27822 is a new contributor. Be nice, and check out our Code of Conduct.
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