Return to Answer

Wikipedia gives $0.51 {km \over s}$ or $510 {m \over s}$ escape velocity, so, no, no leaving Ceres by jumping.

Following my earlier calculations, an asteroid of the radius of Ceres would have orbital speed at near-surface orbit of about $336 {m\over s}$, which is way beyond jump strength of anyone as well.

Gravitational acceleration on the Moon is $1.6249 {m\over s^2}$, barely 6 times more.

With 9h4m day long (32640s) and 3061km equator length, the rotation adds only about $94{m\over s}$ to orbital speed of whatever is standing there - $242 {m\over s}$ remain for orbital speed, so still quite a bit out of human's body reach.

Wikipedia gives $0.51 {km \over s}$ or $510 {m \over s}$ escape velocity, so, no, no leaving Ceres by jumping.

Following my earlier calculations, an asteroid of the radius of Ceres would have orbital speed at near-surface orbit of about $336 {m\over s}$, which is way beyond jump strength of anyone as well.

Gravitational acceleration on the Moon is $1.6249 {m\over s^2}$, barely 6 times more.

With 9h4m day long (32640s) and 3061km equator length, the rotation adds only about $94{m\over s}$ to orbital speed of whatever is standing there - $242 {m\over s}$ remain for orbital speed, so still quite a bit out of human's body reach.

Wikipedia gives $0.51 {km \over s}$ or $510 {m \over s}$ escape velocity, so, no, no leaving Ceres by jumping.

Following my earlier calculations, an asteroid of the radius of Ceres would have orbital speed at near-surface orbit of about $336 {m\over s}$, which is way beyond jump strength of anyone as well.

By the way, did the author of the question mean $0.27 {m\over s^2}$ ?

Gravitational acceleration on the Moon is $1.6249 {m\over s^2}$, barely 6 times more.

With 9h4m day long (32640s) and 3061km equator length, the rotation adds only about $94{m\over s}$ to orbital speed of whatever is standing there - $242 {m\over s}$ remain for orbital speed, so still quite a bit out of human's body reach.

Wikipedia gives $0.51 {km \over s}$ or $510 {m \over s}$ escape velocity, so, no, no leaving Ceres by jumping.

Following my earlier calculations, an asteroid of the radius of Ceres would have orbital speed at near-surface orbit of about $336 {m\over s}$, which is way beyond jump strength of anyone as well.

Gravitational acceleration on the Moon is $1.6249 {m\over s^2}$, barely 6 times more.

With 9h4m day long (32640s) and 3061km equator length, the rotation adds only about $94{m\over s}$ to orbital speed of whatever is standing there - $242 {m\over s}$ remain for orbital speed, so still quite a bit out of human's body reach.

Wikipedia gives $0.51 {km \over s}$ or $510 {m \over s}$ escape velocity, so, no, no leaving Ceres by jumping.

Following my earlier calculations, an asteroid of the radius of Ceres would have orbital speed at near-surface orbit of about $336 {m\over s}$, which is way beyond jump strength of anyone as well.

By the way, did the author of the question mean $0.27 {m\over s^2}$ ?

Gravitational acceleration on the Moon is $1.6249 {m\over s^2}$, barely 6 times more.

Wikipedia gives $0.51 {km \over s}$ or $510 {m \over s}$ escape velocity, so, no, no leaving Ceres by jumping.

Following my earlier calculations, an asteroid of the radius of Ceres would have orbital speed at near-surface orbit of about $336 {m\over s}$, which is way beyond jump strength of anyone as well.

By the way, did the author of the question mean $0.27 {m\over s^2}$ ?

Gravitational acceleration on the Moon is $1.6249 {m\over s^2}$, barely 6 times more.

With 9h4m day long (32640s) and 3061km equator length, the rotation adds only about $94{m\over s}$ to orbital speed of whatever is standing there - $242 {m\over s}$ remain for orbital speed, so still quite a bit out of human's body reach.