How to calculate velocity of a spacecraft hitting the earth

rawemotions wrote:The acceleration caused by gravity of earth is 9.8 m/sec^2.

This essentially means for each second a body is within the sphere of earth's gravity, its velocity is increasing at the rate of 9.8 m/sec.

When a spacecraft re-enters atmosphere, even assuming it starts at 1 m/sec, within 1000 seconds, (assuming straight line approach to Earth's surface) its velocity has increased to an astounding 9.8 km/sec and keeps increasing further , unless the velocity is reduced by atmospheric drag.

When it hits the atmosphere, the friction from atmosphere decelerates , causing re-entry heat issues etc. However is there a way to calculate the velocity (approximate ballpark number) with which the spacecraft will hit the earth, taking into account the time taken for it to hit earth surface (OR just before a parachute opens if there is one). How do we factor in the deceleration caused by atmospheric resistance to calculate this number ?

Or may be re-entering space crafts might fire reverse thrusters to slow down ?

Seva Lamberdar wrote:

rawemotions wrote:The acceleration caused by gravity of earth is 9.8 m/sec^2.

This essentially means for each second a body is within the sphere of earth's gravity, its velocity is increasing at the rate of 9.8 m/sec.

When a spacecraft re-enters atmosphere, even assuming it starts at 1 m/sec, within 1000 seconds, (assuming straight line approach to Earth's surface) its velocity has increased to an astounding 9.8 km/sec and keeps increasing further , unless the velocity is reduced by atmospheric drag.

When it hits the atmosphere, the friction from atmosphere decelerates , causing re-entry heat issues etc. However is there a way to calculate the  velocity (approximate ballpark number) with which the spacecraft will hit the earth, taking into account the time taken for it to hit earth surface (OR just before a parachute opens if there is one). How do we factor in the deceleration caused by atmospheric resistance to calculate this number ?

Or may be  re-entering space crafts might fire reverse thrusters to slow down ?

the velocity and distance of an object (including a spacecraft) traveling through the atmosphere can be calculated by using  a fixed reference (a star etc.). These days the GPS can also be used for this purpose.

rawemotions wrote:

Seva Lamberdar wrote:

rawemotions wrote:The acceleration caused by gravity of earth is 9.8 m/sec^2.

This essentially means for each second a body is within the sphere of earth's gravity, its velocity is increasing at the rate of 9.8 m/sec.

When a spacecraft re-enters atmosphere, even assuming it starts at 1 m/sec, within 1000 seconds, (assuming straight line approach to Earth's surface) its velocity has increased to an astounding 9.8 km/sec and keeps increasing further , unless the velocity is reduced by atmospheric drag.

When it hits the atmosphere, the friction from atmosphere decelerates , causing re-entry heat issues etc. However is there a way to calculate the  velocity (approximate ballpark number) with which the spacecraft will hit the earth, taking into account the time taken for it to hit earth surface (OR just before a parachute opens if there is one). How do we factor in the deceleration caused by atmospheric resistance to calculate this number ?

Or may be  re-entering space crafts might fire reverse thrusters to slow down ?

the velocity and distance of an object (including a spacecraft) traveling through the atmosphere can be calculated by using  a fixed reference (a star etc.). These days the GPS can also be used for this purpose.

It is not so much as the absolute measurement that I am concerned about. Let us assume the object was not moving at all and it wanders into Earth orbit. It is now relative to earth at a Zero velocity. Earth's force of gravity adds a fixed accelaration to that object of ~9.81 m/sec.

My Main question is this. How do we factor in the deceleration caused by atmospheric resistance to calculate the velocity with which the object of mass M will hit the ground ? In other words is there a value for the friction co-efficient (that denotes the opposing fore) caused by the re-entry and further as it travels through the atmosphere to reach earth's surface ? I presume this friction co-efficient would be different for Exosphere, Ionosphere, Stratosphere, Upper and Lower Atmosphere.

Seva Lamberdar wrote:

rawemotions wrote:

Seva Lamberdar wrote:

rawemotions wrote:The acceleration caused by gravity of earth is 9.8 m/sec^2.

This essentially means for each second a body is within the sphere of earth's gravity, its velocity is increasing at the rate of 9.8 m/sec.

When a spacecraft re-enters atmosphere, even assuming it starts at 1 m/sec, within 1000 seconds, (assuming straight line approach to Earth's surface) its velocity has increased to an astounding 9.8 km/sec and keeps increasing further , unless the velocity is reduced by atmospheric drag.

When it hits the atmosphere, the friction from atmosphere decelerates , causing re-entry heat issues etc. However is there a way to calculate the  velocity (approximate ballpark number) with which the spacecraft will hit the earth, taking into account the time taken for it to hit earth surface (OR just before a parachute opens if there is one). How do we factor in the deceleration caused by atmospheric resistance to calculate this number ?

Or may be  re-entering space crafts might fire reverse thrusters to slow down ?

the velocity and distance of an object (including a spacecraft) traveling through the atmosphere can be calculated by using  a fixed reference (a star etc.). These days the GPS can also be used for this purpose.

It is not so much as the absolute measurement that I am concerned about. Let us assume the object was not moving at all and it wanders into Earth orbit. It is now relative to earth at a Zero velocity. Earth's force of gravity adds a fixed accelaration to that object of ~9.81 m/sec.

My Main question is this. How do we factor in the deceleration caused by atmospheric resistance to calculate the velocity with which the object of mass M will hit the ground ? In other words is there a value for the friction co-efficient (that denotes the opposing fore) caused by the re-entry and further as it travels through the atmosphere to reach earth's surface ? I presume this friction co-efficient would be different for Exosphere, Ionosphere, Stratosphere, Upper and Lower Atmosphere.

Of course, the drag force on the object is (inversely) proportional to the square of velocity. Moreover, the constant of proportionality for drag force depends on the atmospheric conditions (density etc., different in lower and upper atmospheres) and the shape and area etc. of the object traveling through the atmosphere.