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First of all, let's figure out what the drag actually is. For that, Heavens-above has a nice chart.

Of some note is the fact that the atmospheric drag rate changes over time, most notably with the solar cycle, but it can change for a variety of reasons, especially for a body as dynamic as the ISS. With the current altitude ranging around 400. The time that it took to go from 414 to 406 km was about 2.5 months, or, say, 75 days. That means that the drag on the spacecraft is about 106 m/day. Orbital energy can be calculated by $e_k=m * v^2/2$. The energy at 406km is 29400301 J, and at 406.1 is 29399868 J*mass. Thus, 433 J*mass of energy is lost per day. The force is applied over all that time period to make that energy lost. $F=m*A$ provides us that 433 J*mass / (Distance traveled in a day) = m*a. Thus, the constant acceleration that would keep it at the same orbit is about $6.56e-7 m/s^2$, or $0.656 \mu m/s^2$.

Given the station mass of 419455kg, the decelerating force would be 0.275 newtons.

First of all, let's figure out what the drag actually is. For that, Heavens-above has a nice chart.

Of some note is the fact that the atmospheric drag rate changes over time, most notably with the solar cycle, but it can change for a variety of reasons, especially for a body as dynamic as the ISS. With the current altitude ranging around 400. The time that it took to go from 414 to 406 km was about 2.5 months, or, say, 75 days. That means that the drag on the spacecraft is about 106 m/day. Orbital energy can be calculated by $e_k=m \cdot v^2/2$. The energy at 406km is 29400301 J/kg, and at 406.1 is 29399868 J/kg. Thus, 433 J/kg of energy is lost per day. The force is applied over all that time period to make that energy lost. $F=m\cdot A$ provides us that 433 J/kg / (Distance traveled in a day) = m * a. Thus, the constant acceleration that would keep it at the same orbit is about $6.56\cdot 10^{-7}~\rm m/s^2$, or $0.656 ~\rm \mu m/s^2$.

Given the station mass of 419455kg, the decelerating force would be 0.275 newtons.

First of all, let's figure out what the drag actually is. For that, Heavens-above has a nice chart.

Of some note is the fact that the atmospheric drag rate changes over time. With the current altitude ranging around 400. The time that it took to go from 414 to 406 km was about 2.5 months, or, say, 75 days. That means that the drag on the spacecraft is about 106 m/day. Orbital energy can be calculated by $e_k=m * v^2/2$. The energy at 406km is 29400301 J, and at 406.1 is 29399868 J*mass. Thus, 433 J*mass of energy is lost per day. The force is applied over all that time period to make that energy lost. $F=m*A$ provides us that 433 J*mass / (Distance traveled in a day) = m*a. Thus, the constant acceleration that would keep it at the same orbit is about $6.56e-7 m/s^2$, or $0.656 \mu m/s^2$.

Given the station mass of 419455kg, the decelerating force would be 0.275 newtons.

First of all, let's figure out what the drag actually is. For that, Heavens-above has a nice chart.

Of some note is the fact that the atmospheric drag rate changes over time, most notably with the solar cycle, but it can change for a variety of reasons, especially for a body as dynamic as the ISS. With the current altitude ranging around 400. The time that it took to go from 414 to 406 km was about 2.5 months, or, say, 75 days. That means that the drag on the spacecraft is about 106 m/day. Orbital energy can be calculated by $e_k=m * v^2/2$. The energy at 406km is 29400301 J, and at 406.1 is 29399868 J*mass. Thus, 433 J*mass of energy is lost per day. The force is applied over all that time period to make that energy lost. $F=m*A$ provides us that 433 J*mass / (Distance traveled in a day) = m*a. Thus, the constant acceleration that would keep it at the same orbit is about $6.56e-7 m/s^2$, or $0.656 \mu m/s^2$.

Given the station mass of 419455kg, the decelerating force would be 0.275 newtons.

First of all, let's figure out what the drag actually is. For that, Heavens-above has a nice chart.

Of some note is the fact that the atmospheric drag rate changes over time. With the current altitude ranging around 400. The time that it took to go from 414 to 406 km was about 2.5 months, or, say, 75 days. That means that the drag on the spacecraft is about 106 m/day. Orbital energy can be calculated by $e_k=m * v^2/2$. The energy at 406km is 29400301 J, and at 406.1 is 29399868 J*mass. Thus, 433 J*mass of energy is lost per day. The force is applied over all that time period to make that energy lost. $F=m*A$ provides us that 433 J*mass / (Distance traveled in a day) = m*a. Thus, the constant acceleration that would keep it at the same orbit is about $6.56e-7 m/s^2$, or $0.656 \mu m/s^2$.

First of all, let's figure out what the drag actually is. For that, Heavens-above has a nice chart.

Of some note is the fact that the atmospheric drag rate changes over time. With the current altitude ranging around 400. The time that it took to go from 414 to 406 km was about 2.5 months, or, say, 75 days. That means that the drag on the spacecraft is about 106 m/day. Orbital energy can be calculated by $e_k=m * v^2/2$. The energy at 406km is 29400301 J, and at 406.1 is 29399868 J*mass. Thus, 433 J*mass of energy is lost per day. The force is applied over all that time period to make that energy lost. $F=m*A$ provides us that 433 J*mass / (Distance traveled in a day) = m*a. Thus, the constant acceleration that would keep it at the same orbit is about $6.56e-7 m/s^2$, or $0.656 \mu m/s^2$.

Given the station mass of 419455kg, the decelerating force would be 0.275 newtons.