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Clarification on Formula of Escape Velocity
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What is R? from the escape velocity formula
$v_e = sqrt frac 2GMR$
Some sources say it is the distance between two objects with mass M and m.
Some examples I have read, only used radius of the M. For simplicity sake, lets use a rocket and planet Earth.
I google the escape velocity of earth as 11.186 km/s. I looked through example such as this. This particular example said that the escape velocity is
$v_e = sqrt 2gR $
$v_e = sqrt 2*9.8*6.4e6 $
$v_e = 11.2 km/s $
I do know that $g = frac GMR^2$, but in this example, the escape velocity is calculated using r as radius of earth. The radius of the Earth according to google is 6371 km or 6.4e6 (the value used for the calculation).
My question is, R the radius of of the earth(M) or the distance between the rocket(m) and Earth? What I think makes sense is the radius of the Earth(M) because how could escape velocity be calculated without knowing the height it has to travel.
To add about the info Distance from M and m, Some sources say that R is actually
$R = r_e + h_atmosphere$
forces gravity potential escape-velocity
asked 1 hour ago
Zirc
16828
add a comment |Â
up vote
2
down vote
favorite
What is R? from the escape velocity formula
$v_e = sqrt frac 2GMR$
Some sources say it is the distance between two objects with mass M and m.
Some examples I have read, only used radius of the M. For simplicity sake, lets use a rocket and planet Earth.
I google the escape velocity of earth as 11.186 km/s. I looked through example such as this. This particular example said that the escape velocity is
$v_e = sqrt 2gR $
$v_e = sqrt 2*9.8*6.4e6 $
$v_e = 11.2 km/s $
I do know that $g = frac GMR^2$, but in this example, the escape velocity is calculated using r as radius of earth. The radius of the Earth according to google is 6371 km or 6.4e6 (the value used for the calculation).
My question is, R the radius of of the earth(M) or the distance between the rocket(m) and Earth? What I think makes sense is the radius of the Earth(M) because how could escape velocity be calculated without knowing the height it has to travel.
To add about the info Distance from M and m, Some sources say that R is actually
$R = r_e + h_atmosphere$
forces gravity potential escape-velocity
asked 1 hour ago
Zirc
16828
It's the inital distance between the (centers of mass of the) two objects. If the rocket is on the surface of the earth it's equal to the radius of the earth
– chiappette
49 mins ago
Your initial formula for $v_e$ is wrong, it shouldn't have the factor of $m$ in it since that factor should cancel out (the units don't work out if you keep that factor in there). Also, your formula for $g$ is wrong, it needs to be $g=GM/R^2$
– enumaris
4 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
What is R? from the escape velocity formula
$v_e = sqrt frac 2GMR$
Some sources say it is the distance between two objects with mass M and m.
Some examples I have read, only used radius of the M. For simplicity sake, lets use a rocket and planet Earth.
I google the escape velocity of earth as 11.186 km/s. I looked through example such as this. This particular example said that the escape velocity is
$v_e = sqrt 2gR $
$v_e = sqrt 2*9.8*6.4e6 $
$v_e = 11.2 km/s $
I do know that $g = frac GMR^2$, but in this example, the escape velocity is calculated using r as radius of earth. The radius of the Earth according to google is 6371 km or 6.4e6 (the value used for the calculation).
My question is, R the radius of of the earth(M) or the distance between the rocket(m) and Earth? What I think makes sense is the radius of the Earth(M) because how could escape velocity be calculated without knowing the height it has to travel.
To add about the info Distance from M and m, Some sources say that R is actually
$R = r_e + h_atmosphere$
forces gravity potential escape-velocity
asked 1 hour ago
Zirc
16828
What is R? from the escape velocity formula
$v_e = sqrt frac 2GMR$
Some sources say it is the distance between two objects with mass M and m.
Some examples I have read, only used radius of the M. For simplicity sake, lets use a rocket and planet Earth.
I google the escape velocity of earth as 11.186 km/s. I looked through example such as this. This particular example said that the escape velocity is
$v_e = sqrt 2gR $
$v_e = sqrt 2*9.8*6.4e6 $
$v_e = 11.2 km/s $
I do know that $g = frac GMR^2$, but in this example, the escape velocity is calculated using r as radius of earth. The radius of the Earth according to google is 6371 km or 6.4e6 (the value used for the calculation).
My question is, R the radius of of the earth(M) or the distance between the rocket(m) and Earth? What I think makes sense is the radius of the Earth(M) because how could escape velocity be calculated without knowing the height it has to travel.
To add about the info Distance from M and m, Some sources say that R is actually
$R = r_e + h_atmosphere$
forces gravity potential escape-velocity
forces gravity potential escape-velocity
asked 1 hour ago
Zirc
16828
asked 1 hour ago
Zirc
16828
edited 15 secs ago
asked 1 hour ago
Zirc
16828
asked 1 hour ago
Zirc
16828
asked 1 hour ago
Zirc
16828
16828
It's the inital distance between the (centers of mass of the) two objects. If the rocket is on the surface of the earth it's equal to the radius of the earth
– chiappette
49 mins ago
Your initial formula for $v_e$ is wrong, it shouldn't have the factor of $m$ in it since that factor should cancel out (the units don't work out if you keep that factor in there). Also, your formula for $g$ is wrong, it needs to be $g=GM/R^2$
– enumaris
4 mins ago
add a comment |Â
It's the inital distance between the (centers of mass of the) two objects. If the rocket is on the surface of the earth it's equal to the radius of the earth
– chiappette
49 mins ago
Your initial formula for $v_e$ is wrong, it shouldn't have the factor of $m$ in it since that factor should cancel out (the units don't work out if you keep that factor in there). Also, your formula for $g$ is wrong, it needs to be $g=GM/R^2$
– enumaris
4 mins ago
It's the inital distance between the (centers of mass of the) two objects. If the rocket is on the surface of the earth it's equal to the radius of the earth
– chiappette
49 mins ago
It's the inital distance between the (centers of mass of the) two objects. If the rocket is on the surface of the earth it's equal to the radius of the earth
– chiappette
49 mins ago
Your initial formula for $v_e$ is wrong, it shouldn't have the factor of $m$ in it since that factor should cancel out (the units don't work out if you keep that factor in there). Also, your formula for $g$ is wrong, it needs to be $g=GM/R^2$
– enumaris
4 mins ago
Your initial formula for $v_e$ is wrong, it shouldn't have the factor of $m$ in it since that factor should cancel out (the units don't work out if you keep that factor in there). Also, your formula for $g$ is wrong, it needs to be $g=GM/R^2$
– enumaris
4 mins ago
add a comment |Â
5 Answers
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$R$ is the distance between the centres of either object. By centres I mean their centres of mass (the point their gravitational forces "pull" from, so to speak).
$R$ is thus to be accurate the sum of Earth's radius $R_e$, the objects radius $r_o$ (if spherical) and the height $h$ it is located above the ground:
$$R=R_e+h+r_o$$
When having small objects you'll often see the radius of the object radius and distance neglected. A rock's small radius is negligible compared to the ~6400 km of the Earth. Adding maybe 1km or even 10km or 100km to this number makes no practical difference.
$$R=R_e+h+r_oapprox R_e$$
But when calculating escape velocity of e.g. Pluto's moon from Pluto, two objects that are comparable in size, you can't neglect either radius. And when calculating escape velocity of our own Moon from Earth, the ~400 000km distance must certainly be included as well (the sizes are almost negligible compared to this).
answered 41 mins ago
Steeven
24.4k557100
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up vote
1
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It is always the distance between the two objects that accounts for the escape velocity. The escape velocity for earth is calculated from the surface of the earth. On the surface, the escape velocity turns out to be around $11.2 km/s$. $R$ given in the problem is the initial distance between the object and the mass's center which also turns out to be the radius of the Earth as we are launching rockets from the surface of the earth. As we go up the distance between the rocket and earth increases and hence the escape velocity decreases.
answered 47 mins ago
シャシュワト
616
add a comment |Â
up vote
1
down vote
Technically, "R" is the radius between the centre of the Earth (or whatever body you're using) and the rocket. That 11.186km/s is the escape velocity at the surface of the Earth, but a hundred kilometers above the Earth the escape velocity is 11.099km/s.
Assuming that you reach escape velocity then begin to coast, your velocity will be constantly decreasing because the pull of the Earth is still acting on your spacecraft. One million kilometers from Earth your speed may have decreased to 1km/s, but that's okay because at that distance the escape velocity is still 0.89km/s.
Hope that helps.
answered 55 mins ago
Ian Moote
1525
add a comment |Â
up vote
1
down vote
R is fundamentally the distance of the spaceship and the celestial. (Read: the ships distance from the planet's centrum exactly at the time it attains the velocity yielded by the formula) Therefore escape velocity depends on where you start from: to leave Earth launching from its surface is much harder than to escape from GEO.
But, to make "gravity wells" of planets comparable, that is to assign numbers to them, indicating reasonably the difficulty of the escape, we must agree on R. Since rockets usually launch from planetary surfaces, escape velocities are calculated using planetary radius as R.
But, since interplanetary missions often have separate stages for orbit insertion and escape burns (or could in the future start out from orbiting outposts), it is also instructive to give escape velocity starting from low orbit (that is, from the height $R_planet+h_atmosphere$) In this case one has simply to subtract her orbital speed from the escape velocity to attain the $Delta v $ needed to reach interplanetary space
answered 34 mins ago
b.Lorenz
26317
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up vote
0
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In order to really understand what this formula means, let's see how to derive that. Let $m$ be the mass of a body on the surface of Earth and let $R_Earth$ be Earth's radius. Our aim is to calculate the initial velocity the body needs, to "escape" from Earth gravitational field. "Escaping", in this context, means to reach the infinity, where the gravitational field is known to be null. Plus, the escape velocity is defined as the minimum velocity needed, physically this means that our body reach the infinity with no more velocity $v_infty = 0$.
Let's start from energy conservation principle:
$E = frac12mv_0^2-fracM_Earth m GR_Earth$
We've just sad the potential at infity is zero, same for the velocity. Thus we have
$frac12mv_0^2-fracM_Earth m GR_Earth=0$.
Solving this equation for $v_0$ we get:
$v_0=sqrt frac2M_EarthGR_Earth$.
If we do not consider air friction, there are no reasons why we should consider the existence of atmosphere.
answered 18 mins ago
Matteo Campagnoli
11
New contributor
Matteo Campagnoli 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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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
$R$ is the distance between the centres of either object. By centres I mean their centres of mass (the point their gravitational forces "pull" from, so to speak).
$R$ is thus to be accurate the sum of Earth's radius $R_e$, the objects radius $r_o$ (if spherical) and the height $h$ it is located above the ground:
$$R=R_e+h+r_o$$
When having small objects you'll often see the radius of the object radius and distance neglected. A rock's small radius is negligible compared to the ~6400 km of the Earth. Adding maybe 1km or even 10km or 100km to this number makes no practical difference.
$$R=R_e+h+r_oapprox R_e$$
But when calculating escape velocity of e.g. Pluto's moon from Pluto, two objects that are comparable in size, you can't neglect either radius. And when calculating escape velocity of our own Moon from Earth, the ~400 000km distance must certainly be included as well (the sizes are almost negligible compared to this).
answered 41 mins ago
Steeven
24.4k557100
add a comment |Â
up vote
2
down vote
$R$ is the distance between the centres of either object. By centres I mean their centres of mass (the point their gravitational forces "pull" from, so to speak).
$R$ is thus to be accurate the sum of Earth's radius $R_e$, the objects radius $r_o$ (if spherical) and the height $h$ it is located above the ground:
$$R=R_e+h+r_o$$
When having small objects you'll often see the radius of the object radius and distance neglected. A rock's small radius is negligible compared to the ~6400 km of the Earth. Adding maybe 1km or even 10km or 100km to this number makes no practical difference.
$$R=R_e+h+r_oapprox R_e$$
But when calculating escape velocity of e.g. Pluto's moon from Pluto, two objects that are comparable in size, you can't neglect either radius. And when calculating escape velocity of our own Moon from Earth, the ~400 000km distance must certainly be included as well (the sizes are almost negligible compared to this).
answered 41 mins ago
Steeven
24.4k557100
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$R$ is the distance between the centres of either object. By centres I mean their centres of mass (the point their gravitational forces "pull" from, so to speak).
$R$ is thus to be accurate the sum of Earth's radius $R_e$, the objects radius $r_o$ (if spherical) and the height $h$ it is located above the ground:
$$R=R_e+h+r_o$$
When having small objects you'll often see the radius of the object radius and distance neglected. A rock's small radius is negligible compared to the ~6400 km of the Earth. Adding maybe 1km or even 10km or 100km to this number makes no practical difference.
$$R=R_e+h+r_oapprox R_e$$
But when calculating escape velocity of e.g. Pluto's moon from Pluto, two objects that are comparable in size, you can't neglect either radius. And when calculating escape velocity of our own Moon from Earth, the ~400 000km distance must certainly be included as well (the sizes are almost negligible compared to this).
answered 41 mins ago
Steeven
24.4k557100
$R$ is the distance between the centres of either object. By centres I mean their centres of mass (the point their gravitational forces "pull" from, so to speak).
$R$ is thus to be accurate the sum of Earth's radius $R_e$, the objects radius $r_o$ (if spherical) and the height $h$ it is located above the ground:
$$R=R_e+h+r_o$$
When having small objects you'll often see the radius of the object radius and distance neglected. A rock's small radius is negligible compared to the ~6400 km of the Earth. Adding maybe 1km or even 10km or 100km to this number makes no practical difference.
$$R=R_e+h+r_oapprox R_e$$
But when calculating escape velocity of e.g. Pluto's moon from Pluto, two objects that are comparable in size, you can't neglect either radius. And when calculating escape velocity of our own Moon from Earth, the ~400 000km distance must certainly be included as well (the sizes are almost negligible compared to this).
answered 41 mins ago
Steeven
24.4k557100
answered 41 mins ago
Steeven
24.4k557100
answered 41 mins ago
Steeven
24.4k557100
answered 41 mins ago
Steeven
24.4k557100
24.4k557100
add a comment |Â
add a comment |Â
up vote
1
down vote
It is always the distance between the two objects that accounts for the escape velocity. The escape velocity for earth is calculated from the surface of the earth. On the surface, the escape velocity turns out to be around $11.2 km/s$. $R$ given in the problem is the initial distance between the object and the mass's center which also turns out to be the radius of the Earth as we are launching rockets from the surface of the earth. As we go up the distance between the rocket and earth increases and hence the escape velocity decreases.
answered 47 mins ago
シャシュワト
616
add a comment |Â
up vote
1
down vote
It is always the distance between the two objects that accounts for the escape velocity. The escape velocity for earth is calculated from the surface of the earth. On the surface, the escape velocity turns out to be around $11.2 km/s$. $R$ given in the problem is the initial distance between the object and the mass's center which also turns out to be the radius of the Earth as we are launching rockets from the surface of the earth. As we go up the distance between the rocket and earth increases and hence the escape velocity decreases.
answered 47 mins ago
シャシュワト
616
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It is always the distance between the two objects that accounts for the escape velocity. The escape velocity for earth is calculated from the surface of the earth. On the surface, the escape velocity turns out to be around $11.2 km/s$. $R$ given in the problem is the initial distance between the object and the mass's center which also turns out to be the radius of the Earth as we are launching rockets from the surface of the earth. As we go up the distance between the rocket and earth increases and hence the escape velocity decreases.
answered 47 mins ago
シャシュワト
616
It is always the distance between the two objects that accounts for the escape velocity. The escape velocity for earth is calculated from the surface of the earth. On the surface, the escape velocity turns out to be around $11.2 km/s$. $R$ given in the problem is the initial distance between the object and the mass's center which also turns out to be the radius of the Earth as we are launching rockets from the surface of the earth. As we go up the distance between the rocket and earth increases and hence the escape velocity decreases.
answered 47 mins ago
シャシュワト
616
answered 47 mins ago
シャシュワト
616
answered 47 mins ago
シャシュワト
616
answered 47 mins ago
シャシュワト
616
616
add a comment |Â
add a comment |Â
up vote
1
down vote
Technically, "R" is the radius between the centre of the Earth (or whatever body you're using) and the rocket. That 11.186km/s is the escape velocity at the surface of the Earth, but a hundred kilometers above the Earth the escape velocity is 11.099km/s.
Assuming that you reach escape velocity then begin to coast, your velocity will be constantly decreasing because the pull of the Earth is still acting on your spacecraft. One million kilometers from Earth your speed may have decreased to 1km/s, but that's okay because at that distance the escape velocity is still 0.89km/s.
Hope that helps.
answered 55 mins ago
Ian Moote
1525
add a comment |Â
up vote
1
down vote
Technically, "R" is the radius between the centre of the Earth (or whatever body you're using) and the rocket. That 11.186km/s is the escape velocity at the surface of the Earth, but a hundred kilometers above the Earth the escape velocity is 11.099km/s.
Assuming that you reach escape velocity then begin to coast, your velocity will be constantly decreasing because the pull of the Earth is still acting on your spacecraft. One million kilometers from Earth your speed may have decreased to 1km/s, but that's okay because at that distance the escape velocity is still 0.89km/s.
Hope that helps.
answered 55 mins ago
Ian Moote
1525
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Technically, "R" is the radius between the centre of the Earth (or whatever body you're using) and the rocket. That 11.186km/s is the escape velocity at the surface of the Earth, but a hundred kilometers above the Earth the escape velocity is 11.099km/s.
Assuming that you reach escape velocity then begin to coast, your velocity will be constantly decreasing because the pull of the Earth is still acting on your spacecraft. One million kilometers from Earth your speed may have decreased to 1km/s, but that's okay because at that distance the escape velocity is still 0.89km/s.
Hope that helps.
answered 55 mins ago
Ian Moote
1525
Technically, "R" is the radius between the centre of the Earth (or whatever body you're using) and the rocket. That 11.186km/s is the escape velocity at the surface of the Earth, but a hundred kilometers above the Earth the escape velocity is 11.099km/s.
Assuming that you reach escape velocity then begin to coast, your velocity will be constantly decreasing because the pull of the Earth is still acting on your spacecraft. One million kilometers from Earth your speed may have decreased to 1km/s, but that's okay because at that distance the escape velocity is still 0.89km/s.
Hope that helps.
answered 55 mins ago
Ian Moote
1525
edited 42 mins ago
answered 55 mins ago
Ian Moote
1525
answered 55 mins ago
Ian Moote
1525
answered 55 mins ago
Ian Moote
1525
1525
add a comment |Â
add a comment |Â
up vote
1
down vote
R is fundamentally the distance of the spaceship and the celestial. (Read: the ships distance from the planet's centrum exactly at the time it attains the velocity yielded by the formula) Therefore escape velocity depends on where you start from: to leave Earth launching from its surface is much harder than to escape from GEO.
But, to make "gravity wells" of planets comparable, that is to assign numbers to them, indicating reasonably the difficulty of the escape, we must agree on R. Since rockets usually launch from planetary surfaces, escape velocities are calculated using planetary radius as R.
But, since interplanetary missions often have separate stages for orbit insertion and escape burns (or could in the future start out from orbiting outposts), it is also instructive to give escape velocity starting from low orbit (that is, from the height $R_planet+h_atmosphere$) In this case one has simply to subtract her orbital speed from the escape velocity to attain the $Delta v $ needed to reach interplanetary space
answered 34 mins ago
b.Lorenz
26317
add a comment |Â
up vote
1
down vote
R is fundamentally the distance of the spaceship and the celestial. (Read: the ships distance from the planet's centrum exactly at the time it attains the velocity yielded by the formula) Therefore escape velocity depends on where you start from: to leave Earth launching from its surface is much harder than to escape from GEO.
But, to make "gravity wells" of planets comparable, that is to assign numbers to them, indicating reasonably the difficulty of the escape, we must agree on R. Since rockets usually launch from planetary surfaces, escape velocities are calculated using planetary radius as R.
But, since interplanetary missions often have separate stages for orbit insertion and escape burns (or could in the future start out from orbiting outposts), it is also instructive to give escape velocity starting from low orbit (that is, from the height $R_planet+h_atmosphere$) In this case one has simply to subtract her orbital speed from the escape velocity to attain the $Delta v $ needed to reach interplanetary space
answered 34 mins ago
b.Lorenz
26317
add a comment |Â
up vote
1
down vote
up vote
1
down vote
R is fundamentally the distance of the spaceship and the celestial. (Read: the ships distance from the planet's centrum exactly at the time it attains the velocity yielded by the formula) Therefore escape velocity depends on where you start from: to leave Earth launching from its surface is much harder than to escape from GEO.
But, to make "gravity wells" of planets comparable, that is to assign numbers to them, indicating reasonably the difficulty of the escape, we must agree on R. Since rockets usually launch from planetary surfaces, escape velocities are calculated using planetary radius as R.
But, since interplanetary missions often have separate stages for orbit insertion and escape burns (or could in the future start out from orbiting outposts), it is also instructive to give escape velocity starting from low orbit (that is, from the height $R_planet+h_atmosphere$) In this case one has simply to subtract her orbital speed from the escape velocity to attain the $Delta v $ needed to reach interplanetary space
answered 34 mins ago
b.Lorenz
26317
R is fundamentally the distance of the spaceship and the celestial. (Read: the ships distance from the planet's centrum exactly at the time it attains the velocity yielded by the formula) Therefore escape velocity depends on where you start from: to leave Earth launching from its surface is much harder than to escape from GEO.
But, to make "gravity wells" of planets comparable, that is to assign numbers to them, indicating reasonably the difficulty of the escape, we must agree on R. Since rockets usually launch from planetary surfaces, escape velocities are calculated using planetary radius as R.
But, since interplanetary missions often have separate stages for orbit insertion and escape burns (or could in the future start out from orbiting outposts), it is also instructive to give escape velocity starting from low orbit (that is, from the height $R_planet+h_atmosphere$) In this case one has simply to subtract her orbital speed from the escape velocity to attain the $Delta v $ needed to reach interplanetary space
answered 34 mins ago
b.Lorenz
26317
answered 34 mins ago
b.Lorenz
26317
answered 34 mins ago
b.Lorenz
26317
answered 34 mins ago
b.Lorenz
26317
26317
add a comment |Â
add a comment |Â
up vote
0
down vote
In order to really understand what this formula means, let's see how to derive that. Let $m$ be the mass of a body on the surface of Earth and let $R_Earth$ be Earth's radius. Our aim is to calculate the initial velocity the body needs, to "escape" from Earth gravitational field. "Escaping", in this context, means to reach the infinity, where the gravitational field is known to be null. Plus, the escape velocity is defined as the minimum velocity needed, physically this means that our body reach the infinity with no more velocity $v_infty = 0$.
Let's start from energy conservation principle:
$E = frac12mv_0^2-fracM_Earth m GR_Earth$
We've just sad the potential at infity is zero, same for the velocity. Thus we have
$frac12mv_0^2-fracM_Earth m GR_Earth=0$.
Solving this equation for $v_0$ we get:
$v_0=sqrt frac2M_EarthGR_Earth$.
If we do not consider air friction, there are no reasons why we should consider the existence of atmosphere.
answered 18 mins ago
Matteo Campagnoli
11
New contributor
Matteo Campagnoli 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
0
down vote
In order to really understand what this formula means, let's see how to derive that. Let $m$ be the mass of a body on the surface of Earth and let $R_Earth$ be Earth's radius. Our aim is to calculate the initial velocity the body needs, to "escape" from Earth gravitational field. "Escaping", in this context, means to reach the infinity, where the gravitational field is known to be null. Plus, the escape velocity is defined as the minimum velocity needed, physically this means that our body reach the infinity with no more velocity $v_infty = 0$.
Let's start from energy conservation principle:
$E = frac12mv_0^2-fracM_Earth m GR_Earth$
We've just sad the potential at infity is zero, same for the velocity. Thus we have
$frac12mv_0^2-fracM_Earth m GR_Earth=0$.
Solving this equation for $v_0$ we get:
$v_0=sqrt frac2M_EarthGR_Earth$.
If we do not consider air friction, there are no reasons why we should consider the existence of atmosphere.
answered 18 mins ago
Matteo Campagnoli
11
New contributor
Matteo Campagnoli 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
0
down vote
up vote
0
down vote
In order to really understand what this formula means, let's see how to derive that. Let $m$ be the mass of a body on the surface of Earth and let $R_Earth$ be Earth's radius. Our aim is to calculate the initial velocity the body needs, to "escape" from Earth gravitational field. "Escaping", in this context, means to reach the infinity, where the gravitational field is known to be null. Plus, the escape velocity is defined as the minimum velocity needed, physically this means that our body reach the infinity with no more velocity $v_infty = 0$.
Let's start from energy conservation principle:
$E = frac12mv_0^2-fracM_Earth m GR_Earth$
We've just sad the potential at infity is zero, same for the velocity. Thus we have
$frac12mv_0^2-fracM_Earth m GR_Earth=0$.
Solving this equation for $v_0$ we get:
$v_0=sqrt frac2M_EarthGR_Earth$.
If we do not consider air friction, there are no reasons why we should consider the existence of atmosphere.
answered 18 mins ago
Matteo Campagnoli
11
New contributor
Matteo Campagnoli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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In order to really understand what this formula means, let's see how to derive that. Let $m$ be the mass of a body on the surface of Earth and let $R_Earth$ be Earth's radius. Our aim is to calculate the initial velocity the body needs, to "escape" from Earth gravitational field. "Escaping", in this context, means to reach the infinity, where the gravitational field is known to be null. Plus, the escape velocity is defined as the minimum velocity needed, physically this means that our body reach the infinity with no more velocity $v_infty = 0$.
Let's start from energy conservation principle:
$E = frac12mv_0^2-fracM_Earth m GR_Earth$
We've just sad the potential at infity is zero, same for the velocity. Thus we have
$frac12mv_0^2-fracM_Earth m GR_Earth=0$.
Solving this equation for $v_0$ we get:
$v_0=sqrt frac2M_EarthGR_Earth$.
If we do not consider air friction, there are no reasons why we should consider the existence of atmosphere.
answered 18 mins ago
Matteo Campagnoli
11
New contributor
Matteo Campagnoli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 18 mins ago
Matteo Campagnoli
11
New contributor
Matteo Campagnoli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 18 mins ago
Matteo Campagnoli
11
answered 18 mins ago
Matteo Campagnoli
11
11
New contributor
Matteo Campagnoli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Matteo Campagnoli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Matteo Campagnoli 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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It's the inital distance between the (centers of mass of the) two objects. If the rocket is on the surface of the earth it's equal to the radius of the earth
– chiappette
49 mins ago
Your initial formula for $v_e$ is wrong, it shouldn't have the factor of $m$ in it since that factor should cancel out (the units don't work out if you keep that factor in there). Also, your formula for $g$ is wrong, it needs to be $g=GM/R^2$
– enumaris
4 mins ago