How can a rocket approaching the Karman Line then return to earth faster than 53 m/s terminal velocity?
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In the book, How to Make a Spaceship: A Band of Renegades, an Epic Race and the Birth of Private Spaceflight the opening chapter details the June 2004 attempt to take SpaceShipOne to the Karman Line.
The chapter also recounts a story of the X-15, piloted by Mike Adams in 1967, re-entering the earth's atmosphere at Mach 5 and breaking up in a violent spin. This is used as a warning to the reader about the dangers that Melvill is facing in SpaceShipOne.
How does a vehicle, having exhausted it's primary thrust, return to earth at a greater speed than terminal velocity?
rockets gravity thrust
asked 2 hours ago
Venture2099
223127
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up vote
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In the book, How to Make a Spaceship: A Band of Renegades, an Epic Race and the Birth of Private Spaceflight the opening chapter details the June 2004 attempt to take SpaceShipOne to the Karman Line.
The chapter also recounts a story of the X-15, piloted by Mike Adams in 1967, re-entering the earth's atmosphere at Mach 5 and breaking up in a violent spin. This is used as a warning to the reader about the dangers that Melvill is facing in SpaceShipOne.
How does a vehicle, having exhausted it's primary thrust, return to earth at a greater speed than terminal velocity?
rockets gravity thrust
asked 2 hours ago
Venture2099
223127
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In the book, How to Make a Spaceship: A Band of Renegades, an Epic Race and the Birth of Private Spaceflight the opening chapter details the June 2004 attempt to take SpaceShipOne to the Karman Line.
The chapter also recounts a story of the X-15, piloted by Mike Adams in 1967, re-entering the earth's atmosphere at Mach 5 and breaking up in a violent spin. This is used as a warning to the reader about the dangers that Melvill is facing in SpaceShipOne.
How does a vehicle, having exhausted it's primary thrust, return to earth at a greater speed than terminal velocity?
rockets gravity thrust
asked 2 hours ago
Venture2099
223127
In the book, How to Make a Spaceship: A Band of Renegades, an Epic Race and the Birth of Private Spaceflight the opening chapter details the June 2004 attempt to take SpaceShipOne to the Karman Line.
The chapter also recounts a story of the X-15, piloted by Mike Adams in 1967, re-entering the earth's atmosphere at Mach 5 and breaking up in a violent spin. This is used as a warning to the reader about the dangers that Melvill is facing in SpaceShipOne.
How does a vehicle, having exhausted it's primary thrust, return to earth at a greater speed than terminal velocity?
rockets gravity thrust
rockets gravity thrust
asked 2 hours ago
Venture2099
223127
asked 2 hours ago
Venture2099
223127
asked 2 hours ago
Venture2099
223127
asked 2 hours ago
Venture2099
223127
asked 2 hours ago
Venture2099
223127
223127
add a comment |
add a comment |
1 Answer
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53 m/s is the approximate terminal velocity of a human skydiver.
The terminal velocity of a 7-ton metal dart is quite a bit higher. Larger objects tend to be affected less by atmospheric drag than smaller ones, all other things being equal. Terminal velocity also increases with altitude because the air is thinner.
Assuming 7000 kg mass, 3.5 m2 cross section, and coefficient of drag 0.3 (educated guesses for the X-15), this online terminal velocity calculator gives 295 m/s terminal velocity at sea level (about 0.9 mach). At 25km altitude (air density 0.04), terminal velocity is about 1800 m/s or mach 5.
answered 2 hours ago
Russell Borogove
77.3k2248335
I thought all things had the same terminal velocity. If we drop a book and a cannonball from a great height they land at the same time no?
– Venture2099
2 hours ago
2
@Venture2099 Not in atmosphere, no (do a feather and a cannonball land at the same time?). The retarding force from air resistance is proportional to cross-sectional area; the relative acceleration imparted by that force is inversely proportional to mass. For spheres of identical density, for example, cross sectional area goes up by the square of radius; mass goes up by the cube of radius, so drag-acceleration is inversely proportional to radius.
– Russell Borogove
2 hours ago
@Venture2099 The reason a demonstration with two metal balls from not too great a height can work is just that the effect of air resistance will be negligible for fairly massive objects at fairly low velocities.
– Mark Foskey
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
53 m/s is the approximate terminal velocity of a human skydiver.
The terminal velocity of a 7-ton metal dart is quite a bit higher. Larger objects tend to be affected less by atmospheric drag than smaller ones, all other things being equal. Terminal velocity also increases with altitude because the air is thinner.
Assuming 7000 kg mass, 3.5 m2 cross section, and coefficient of drag 0.3 (educated guesses for the X-15), this online terminal velocity calculator gives 295 m/s terminal velocity at sea level (about 0.9 mach). At 25km altitude (air density 0.04), terminal velocity is about 1800 m/s or mach 5.
answered 2 hours ago
Russell Borogove
77.3k2248335
I thought all things had the same terminal velocity. If we drop a book and a cannonball from a great height they land at the same time no?
– Venture2099
2 hours ago
2
@Venture2099 Not in atmosphere, no (do a feather and a cannonball land at the same time?). The retarding force from air resistance is proportional to cross-sectional area; the relative acceleration imparted by that force is inversely proportional to mass. For spheres of identical density, for example, cross sectional area goes up by the square of radius; mass goes up by the cube of radius, so drag-acceleration is inversely proportional to radius.
– Russell Borogove
2 hours ago
@Venture2099 The reason a demonstration with two metal balls from not too great a height can work is just that the effect of air resistance will be negligible for fairly massive objects at fairly low velocities.
– Mark Foskey
1 hour ago
add a comment |
up vote
4
down vote
53 m/s is the approximate terminal velocity of a human skydiver.
The terminal velocity of a 7-ton metal dart is quite a bit higher. Larger objects tend to be affected less by atmospheric drag than smaller ones, all other things being equal. Terminal velocity also increases with altitude because the air is thinner.
Assuming 7000 kg mass, 3.5 m2 cross section, and coefficient of drag 0.3 (educated guesses for the X-15), this online terminal velocity calculator gives 295 m/s terminal velocity at sea level (about 0.9 mach). At 25km altitude (air density 0.04), terminal velocity is about 1800 m/s or mach 5.
answered 2 hours ago
Russell Borogove
77.3k2248335
I thought all things had the same terminal velocity. If we drop a book and a cannonball from a great height they land at the same time no?
– Venture2099
2 hours ago
2
@Venture2099 Not in atmosphere, no (do a feather and a cannonball land at the same time?). The retarding force from air resistance is proportional to cross-sectional area; the relative acceleration imparted by that force is inversely proportional to mass. For spheres of identical density, for example, cross sectional area goes up by the square of radius; mass goes up by the cube of radius, so drag-acceleration is inversely proportional to radius.
– Russell Borogove
2 hours ago
@Venture2099 The reason a demonstration with two metal balls from not too great a height can work is just that the effect of air resistance will be negligible for fairly massive objects at fairly low velocities.
– Mark Foskey
1 hour ago
add a comment |
up vote
4
down vote
up vote
4
down vote
53 m/s is the approximate terminal velocity of a human skydiver.
The terminal velocity of a 7-ton metal dart is quite a bit higher. Larger objects tend to be affected less by atmospheric drag than smaller ones, all other things being equal. Terminal velocity also increases with altitude because the air is thinner.
Assuming 7000 kg mass, 3.5 m2 cross section, and coefficient of drag 0.3 (educated guesses for the X-15), this online terminal velocity calculator gives 295 m/s terminal velocity at sea level (about 0.9 mach). At 25km altitude (air density 0.04), terminal velocity is about 1800 m/s or mach 5.
answered 2 hours ago
Russell Borogove
77.3k2248335
53 m/s is the approximate terminal velocity of a human skydiver.
The terminal velocity of a 7-ton metal dart is quite a bit higher. Larger objects tend to be affected less by atmospheric drag than smaller ones, all other things being equal. Terminal velocity also increases with altitude because the air is thinner.
Assuming 7000 kg mass, 3.5 m2 cross section, and coefficient of drag 0.3 (educated guesses for the X-15), this online terminal velocity calculator gives 295 m/s terminal velocity at sea level (about 0.9 mach). At 25km altitude (air density 0.04), terminal velocity is about 1800 m/s or mach 5.
answered 2 hours ago
Russell Borogove
77.3k2248335
edited 1 hour ago
answered 2 hours ago
Russell Borogove
77.3k2248335
answered 2 hours ago
Russell Borogove
77.3k2248335
answered 2 hours ago
Russell Borogove
77.3k2248335
77.3k2248335
I thought all things had the same terminal velocity. If we drop a book and a cannonball from a great height they land at the same time no?
– Venture2099
2 hours ago
2
@Venture2099 Not in atmosphere, no (do a feather and a cannonball land at the same time?). The retarding force from air resistance is proportional to cross-sectional area; the relative acceleration imparted by that force is inversely proportional to mass. For spheres of identical density, for example, cross sectional area goes up by the square of radius; mass goes up by the cube of radius, so drag-acceleration is inversely proportional to radius.
– Russell Borogove
2 hours ago
@Venture2099 The reason a demonstration with two metal balls from not too great a height can work is just that the effect of air resistance will be negligible for fairly massive objects at fairly low velocities.
– Mark Foskey
1 hour ago
add a comment |
I thought all things had the same terminal velocity. If we drop a book and a cannonball from a great height they land at the same time no?
– Venture2099
2 hours ago
2
@Venture2099 Not in atmosphere, no (do a feather and a cannonball land at the same time?). The retarding force from air resistance is proportional to cross-sectional area; the relative acceleration imparted by that force is inversely proportional to mass. For spheres of identical density, for example, cross sectional area goes up by the square of radius; mass goes up by the cube of radius, so drag-acceleration is inversely proportional to radius.
– Russell Borogove
2 hours ago
@Venture2099 The reason a demonstration with two metal balls from not too great a height can work is just that the effect of air resistance will be negligible for fairly massive objects at fairly low velocities.
– Mark Foskey
1 hour ago
I thought all things had the same terminal velocity. If we drop a book and a cannonball from a great height they land at the same time no?
– Venture2099
2 hours ago
I thought all things had the same terminal velocity. If we drop a book and a cannonball from a great height they land at the same time no?
– Venture2099
2 hours ago
2
2
@Venture2099 Not in atmosphere, no (do a feather and a cannonball land at the same time?). The retarding force from air resistance is proportional to cross-sectional area; the relative acceleration imparted by that force is inversely proportional to mass. For spheres of identical density, for example, cross sectional area goes up by the square of radius; mass goes up by the cube of radius, so drag-acceleration is inversely proportional to radius.
– Russell Borogove
2 hours ago
@Venture2099 Not in atmosphere, no (do a feather and a cannonball land at the same time?). The retarding force from air resistance is proportional to cross-sectional area; the relative acceleration imparted by that force is inversely proportional to mass. For spheres of identical density, for example, cross sectional area goes up by the square of radius; mass goes up by the cube of radius, so drag-acceleration is inversely proportional to radius.
– Russell Borogove
2 hours ago
@Venture2099 The reason a demonstration with two metal balls from not too great a height can work is just that the effect of air resistance will be negligible for fairly massive objects at fairly low velocities.
– Mark Foskey
1 hour ago
@Venture2099 The reason a demonstration with two metal balls from not too great a height can work is just that the effect of air resistance will be negligible for fairly massive objects at fairly low velocities.
– Mark Foskey
1 hour ago
add a comment |
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