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Gravitational force
Storyboard 
Gravitational force is defined as the gravitational mass multiplied by the gravitational acceleration.
The gravitational acceleration depends on the planet or moon being considered. While on Earth, the gravitational acceleration $g$ is 9.8 m/s², on the Moon, it's 1.625 m/s².
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Equations
$ F = F_g $
F = F_g
$ F = m_i a_0 $
F = m_i * a
$ F_g = m_g g $
F_g = m_g * g
$ m_g = m_i $
m_g = m_i
$ s = s_0 + v_0 ( t - t_0 )+\displaystyle\frac{1}{2} a_0 ( t - t_0 )^2$
s = s_0 + v_0 * ( t - t_0 )+ a_0 *( t - t_0 )^2/2
$ s = s_0 +\displaystyle\frac{ v ^2- v_0 ^2}{2 a_0 }$
s = s_0 +( v ^2- v_0 ^2)/(2* a_0 )
$ v = v_0 + a_0 ( t - t_0 )$
v = v_0 + a_0 *( t - t_0 )
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Gravitational mass concept
Concept
Gravitational mass is associated with what Newton defined as the law of gravitation, indicating the force that one body exerts on another.
It should not be confused with inertial mass, which indicates the resistance a body generates when changing its state of motion. The latter is associated with the inertia experienced by bodies and is referred to as inertial mass.
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Inertial and gravitational mass equality
Video
The gravitational mass
This was shown by the astronauts in Apollo 15. The first part contains the original video, the second a Hollywood type version.
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Equality of inertial and gravitational mass
Equation
The masses that Newton used in his principles are related to the inertia of bodies, which leads to the concept of the inertial Mass ($m_i$).
Newton's law, which is linked to the force between bodies due to their masses, is related to gravity, hence known as the gravitational mass ($m_g$).
Empirically, it has been concluded that both masses are equivalent, and therefore, we define
| $ m_g = m_i $ |
Einstein was the one who questioned this equality and, from that doubt, understood why both 'appear' equal in his theory of gravity. In his argument, Einstein explained that masses deform space, and this deformation of space causes a change in the behavior of bodies. Thus, masses turn out to be equivalent. The revolutionary concept of space curvature implies that even light, which lacks mass, is affected by celestial bodies, contradicting Newton's theory of gravitation. This was experimentally demonstrated by studying the behavior of light during a solar eclipse. In this situation, light beams are deflected due to the presence of the sun, allowing stars behind it to be observed.
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Acceleration in gravitational field
Equation
When a force is applied to a mass, propelling it within the Earth's gravitational field, the following relationship arises:
| $ F = F_g $ |
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Force case constant mass
Equation
In the case where the inertial Mass ($m_i$) equals the initial mass ($m_0$),
| $ m_g = m_i $ |
the derivative of momentum will be equal to the mass multiplied by the derivative of the speed ($v$). Since the derivative of velocity is the instant acceleration ($a$), we have that the force with constant mass ($F$) is
| $ F = m_i a_0 $ |
| $ F = m_i a $ |
Since the moment ($p$) is defined with the inertial Mass ($m_i$) and the speed ($v$),
| $ p = m_i v $ |
If the inertial Mass ($m_i$) is equal to the initial mass ($m_0$), then we can derive the momentum with respect to time and obtain the force with constant mass ($F$):
$F=\displaystyle\frac{d}{dt}p=m_i\displaystyle\frac{d}{dt}v=m_ia$
Therefore, we conclude that
| $ F = m_i a $ |
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Gravitational Force
Equation
The gravitational Force ($F_g$) is based on the gravitational mass ($m_g$) of the object and on a constant reflecting the intensity of gravity at the planet's surface. The latter is identified by the gravitational Acceleration ($g$), which is equal to $9.8 m/s^2$.
Consequently, it is concluded that:
| $ F_g = m_g g $ |
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Speed by constant acceleration
Equation
If the constant Acceleration ($a_0$), then the mean Acceleration ($\bar{a}$) is equal to the value of acceleration, that is,
| $ a_0 = \bar{a} $ |
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In this case, the speed ($v$) as a function of the time ($t$) can be calculated by considering that it is associated with the difference between the speed ($v$) and the initial Speed ($v_0$), as well as the time ($t$) and the start Time ($t_0$).
| $ v = v_0 + a_0 ( t - t_0 )$ |
In the case where the constant Acceleration ($a_0$) equals the mean Acceleration ($\bar{a}$), it will be equal to
| $ a_0 = \bar{a} $ |
.
Therefore, considering the speed Diference ($\Delta v$) as
| $ \Delta v \equiv v - v_0 $ |
and the time elapsed ($\Delta t$) as
| $ \Delta t \equiv t - t_0 $ |
,
the equation for the constant Acceleration ($a_0$)
| $ \bar{a} \equiv\displaystyle\frac{ \Delta v }{ \Delta t }$ |
can be written as
$a_0 = \bar{a} = \displaystyle\frac{\Delta v}{\Delta t} = \displaystyle\frac{v - v_0}{t - t_0}$
and by rearranging, we obtain
| $ v = v_0 + a_0 ( t - t_0 )$ |
.
This equation thus represents a straight line in velocity-time space.
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Path traveled with constant acceleration
Equation
In the case of ($$), the speed ($v$) varies linearly with the time ($t$), using the initial Speed ($v_0$) and the start Time ($t_0$):
| $ v = v_0 + a_0 ( t - t_0 )$ |
Thus, the area under this line can be calculated, yielding the distance traveled in a time ($\Delta s$). Combining this with the starting position ($s_0$), we can calculate the position ($s$), resulting in:
| $ s = s_0 + v_0 ( t - t_0 )+\displaystyle\frac{1}{2} a_0 ( t - t_0 )^2$ |
In the case of the constant Acceleration ($a_0$), the speed ($v$) as a function of the time ($t$) forms a straight line passing through the start Time ($t_0$) and the initial Speed ($v_0$), defined by the equation:
| $ v = v_0 + a_0 ( t - t_0 )$ |
Since the distance traveled in a time ($\Delta s$) represents the area under the velocity-time curve, we can sum the contributions of the rectangle:
$v_0(t-t_0)$
and the triangle:
$\displaystyle\frac{1}{2}a_0(t-t_0)^2$
To obtain the distance traveled in a time ($\Delta s$) with the position ($s$) and the starting position ($s_0$), resulting in:
| $ \Delta s \equiv s - s_0 $ |
Therefore:
| $ s = s_0 + v_0 ( t - t_0 )+\displaystyle\frac{1}{2} a_0 ( t - t_0 )^2$ |
This corresponds to the general form of a parabola.
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Path with Constant Acceleration as a Function of the Speed
Equation
In the case of constant acceleration, we can calculate the position ($s$) from the starting position ($s_0$), the initial Speed ($v_0$), the time ($t$), and the start Time ($t_0$) using the equation:
| $ s = s_0 + v_0 ( t - t_0 )+\displaystyle\frac{1}{2} a_0 ( t - t_0 )^2$ |
This allows us to determine the relationship between the distance covered during acceleration/deceleration and the change in velocity:
| $ s = s_0 +\displaystyle\frac{ v ^2- v_0 ^2}{2 a_0 }$ |
If we solve for the time ($t$) and the start Time ($t_0$) in the equation of the speed ($v$), which depends on the initial Speed ($v_0$) and the constant Acceleration ($a_0$):
| $ v = v_0 + a_0 ( t - t_0 )$ |
we get:
$t - t_0= \displaystyle\frac{v - v_0}{a_0}$
And when we substitute this into the equation of the position ($s$) with the starting position ($s_0$):
| $ s = s_0 + v_0 ( t - t_0 )+\displaystyle\frac{1}{2} a_0 ( t - t_0 )^2$ |
we obtain an expression for the distance traveled as a function of velocity:
| $ s = s_0 +\displaystyle\frac{ v ^2- v_0 ^2}{2 a_0 }$ |
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