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Is the gravitational force less? Yes. By how much? By an insignificant amount.

The gravitational force of attraction between two objects is given by,

$\displaystyle F_{\mathrm g} = \frac{G m_{1} m_{2}}{R^2}$,

where,

$G$ is the graviational constant,

$R$ is the distance between the object's centers, and

$m_{1}$ and $m_{2}$ are the masses of the objects.

Instead of finding the variation in force between the aircraft and earth, it would be be better to find the variation in the acceleration due to gravity, $g$ (as $F_{\mathrm g} = m_{\mathrm a} g$, with $m_{\mathrm a}$ being the mass of the airliner)

We have, on earth's surface,

$\displaystyle g = \frac{G m_{\mathrm e}}{R_{\mathrm e}^2}$

where,

$m_{\mathrm e}$ is the mass of the earth, and

$R_{\mathrm e}$ is the radius of the earth.

For the aircraft at an altitude $h$ above the surface of the earth, this becomes,

$\displaystyle g_{h} = \frac{G m_{\mathrm e}}{\left(R_{\mathrm e} + h\right)^2}$

Taking ratio, we get,

$\displaystyle \frac{g_{h}}{g} = \left(1 + \frac{h}{R_{e}}\right)^{-2}$

Plugging in numbers, we get, for an airliner cruising at 12 km,

$g_{h} = 9.773\ \mathrm{m\ s^{-2}}$,

or about 0.37 % less compared to the sea level value.

  This is quite small and would not be noticable to all but the sensitive instruments.

There is less gravitational force, but by how much? An insignificant amount. The gravitational force of attraction between two objects is given by,

$\displaystyle F_{\mathrm g} = \frac{G m_{1} m_{2}}{R^2}$,

where,

$G$ is the graviational constant,

$R$ is the distance between the object's centers, and

$m_{1}$ and $m_{2}$ are the masses of the objects.

Instead of finding the variation in force between the aircraft and earth, it would be be better to find the variation in the acceleration due to gravity, $g$ (as $F_{\mathrm g} = m_{\mathrm a} g$, with $m_{\mathrm a}$ being the mass of the airliner)

We have, on earth's surface,

$\displaystyle g = \frac{G m_{\mathrm e}}{R_{\mathrm e}^2}$

where,

$m_{\mathrm e}$ is the mass of the earth, and

$R_{\mathrm e}$ is the radius of the earth.

For the aircraft at an altitude $h$ above the surface of the earth, this becomes,

$\displaystyle g_{h} = \frac{G m_{\mathrm e}}{\left(R_{\mathrm e} + h\right)^2}$

Taking ratio, we get,

$\displaystyle \frac{g_{h}}{g} = \left(1 + \frac{h}{R_{e}}\right)^{-2}$

Plugging in numbers, we get, for an airliner cruising at 12 km,

$g_{h} = 9.773\ \mathrm{m\ s^{-2}}$,

or about 0.37 % less compared to the sea level value. This is quite small and would not be noticable to all but the sensitive instruments.

Is the gravitational force less? Yes. By how much? By an insignificant amount.

The gravitational force of attration between two objects is given by,

$\displaystyle F_{\mathrm g} = \frac{G m_{1} m_{2}}{R^2}$,

where,

$G$ is the graviational constant,

$R$ is the distance between the object's centers, and

$m_{1}$ and $m_{2}$ are the masses of the objects.

Instead of finding the variation in force between the aircraft and earth, it wouldbe be better to find the variation in the acceleration due to gravity, $g$ (as $F_{\mathrm g} = m_{\mathrm a} g$, with $m_{\mathrm a}$ being the mass of the airliner)

We have, on earth's surface,

$\displaystyle g = \frac{G m_{\mathrm e}}{R_{\mathrm e}^2}$

where,

$m_{\mathrm e}$ is the mass of the earth, and

$R_{\mathrm e}$ is the radius of the earth.

For the aircraft at an altitude $h$ above the surface of the earth, this becomes,

$\displaystyle g_{h} = \frac{G m_{\mathrm e}}{\left(R_{\mathrm e} + h\right)^2}$

Taking ratio, we get,

$\displaystyle \frac{g_{h}}{g} = \left(1 + \frac{h}{R_{e}}\right)^{-2}$

Plugging in numbers, we get, for an airliner cruising at 12 km,

$g_{h} = 9.773\ \mathrm{m\ s^{-2}}$,

or about 0.37 % less compared to the sea level value.

This is quite small and would not be noticable to all but the sensitive instruments.

Is the gravitational force less? Yes. By how much? By an insignificant amount.

The gravitational force of attraction between two objects is given by,

$\displaystyle F_{\mathrm g} = \frac{G m_{1} m_{2}}{R^2}$,

where,

$G$ is the graviational constant,

$R$ is the distance between the object's centers, and

$m_{1}$ and $m_{2}$ are the masses of the objects.

Instead of finding the variation in force between the aircraft and earth, it would be be better to find the variation in the acceleration due to gravity, $g$ (as $F_{\mathrm g} = m_{\mathrm a} g$, with $m_{\mathrm a}$ being the mass of the airliner)

We have, on earth's surface,

$\displaystyle g = \frac{G m_{\mathrm e}}{R_{\mathrm e}^2}$

where,

$m_{\mathrm e}$ is the mass of the earth, and

$R_{\mathrm e}$ is the radius of the earth.

For the aircraft at an altitude $h$ above the surface of the earth, this becomes,

$\displaystyle g_{h} = \frac{G m_{\mathrm e}}{\left(R_{\mathrm e} + h\right)^2}$

Taking ratio, we get,

$\displaystyle \frac{g_{h}}{g} = \left(1 + \frac{h}{R_{e}}\right)^{-2}$

Plugging in numbers, we get, for an airliner cruising at 12 km,

$g_{h} = 9.773\ \mathrm{m\ s^{-2}}$,

or about 0.37 % less compared to the sea level value.

This is quite small and would not be noticable to all but the sensitive instruments.

Is the gravitational force less? Yes. By how much? By an insignificant amount.

The gravitational force of attration between two objects is given by,

$\displaystyle F_{\mathrm g} = \frac{G m_{1} m_{2}}{R^2}$,

where,

$G$ is the graviational constant,

$R$ is the distance between the object's centers, and

$m_{1}$ and $m_{2}$ are the masses of the objects.

Instead of finding the variation in force between the aircraft and earth, it wouldbe be better to find the variation in the acceleration due to gravity, $g$ (as $F_{\mathrm g} = m_{\mathrm a} g$, with $m_{\mathrm a}$ being the mass of the airliner)

We have, on earth's surface,

$\displaystyle g = \frac{G m_{\mathrm e}}{R_{\mathrm e}^2}$

where,

$m_{\mathrm e}$ is the mass of the earth, and

$R_{\mathrm e}$ is the radius of the earth.

For the aircraft at an altitude $h$ above the surface of the earth, this becomes,

$\displaystyle g_{h} = \frac{G m_{\mathrm e}}{\left(R_{\mathrm e} + h\right)^2}$

Taking ratio, we get,

$\displaystyle \frac{g_{h}}{g} = \left(1 + \frac{h}{R_{e}}\right)^{-2}$

Plugging in numbers, we get, for an airliner cruising at 12 km,

$g_{h} = 9.773\ \mathrm{m\ s^{-2}}$,

or about 0.37 % less compared to the sea level value.

This is quite less and would not be noticable to all but the sensitive instruments.

Is the gravitational force less? Yes. By how much? By an insignificant amount.

The gravitational force of attration between two objects is given by,

$\displaystyle F_{\mathrm g} = \frac{G m_{1} m_{2}}{R^2}$,

where,

$G$ is the graviational constant,

$R$ is the distance between the object's centers, and

$m_{1}$ and $m_{2}$ are the masses of the objects.

Instead of finding the variation in force between the aircraft and earth, it wouldbe be better to find the variation in the acceleration due to gravity, $g$ (as $F_{\mathrm g} = m_{\mathrm a} g$, with $m_{\mathrm a}$ being the mass of the airliner)

We have, on earth's surface,

$\displaystyle g = \frac{G m_{\mathrm e}}{R_{\mathrm e}^2}$

where,

$m_{\mathrm e}$ is the mass of the earth, and

$R_{\mathrm e}$ is the radius of the earth.

For the aircraft at an altitude $h$ above the surface of the earth, this becomes,

$\displaystyle g_{h} = \frac{G m_{\mathrm e}}{\left(R_{\mathrm e} + h\right)^2}$

Taking ratio, we get,

$\displaystyle \frac{g_{h}}{g} = \left(1 + \frac{h}{R_{e}}\right)^{-2}$

Plugging in numbers, we get, for an airliner cruising at 12 km,

$g_{h} = 9.773\ \mathrm{m\ s^{-2}}$,

or about 0.37 % less compared to the sea level value.

This is quite small and would not be noticable to all but the sensitive instruments.