# gravitational acceleration

#### dwsmith

##### Well-known member
A satellite 300km above the earth's radius would have the same gravitational acceleration magnitude?
$$\mathbf{F} = \frac{GM_{e}m_s}{r^2}\approx \frac{GM_e}{r^2}$$
Correct?

#### Ackbach

##### Indicium Physicus
Staff member
I would write that

$$\mathbf{F}_{es}=-G\frac{M_{e}M_{s}}{|\mathbf{r}_{es}|^{2}}\, \hat{\mathbf{r}}_{es}.$$

Here $\mathbf{F}_{es}$ is the force exerted on the satellite due to the earth, and $\mathbf{r}_{es}$ is the radius vector from the earth to the satellite. Therefore, by Newton's Second Law, we have that
$$\mathbf{F}_{es}=M_{s}\mathbf{a}_{s},$$
and hence
$$\mathbf{a}_{s}=-G\frac{M_{e}}{|\mathbf{r}_{es}|^{2}}\, \hat{\mathbf{r}}_{es}.$$

#### dwsmith

##### Well-known member
I would write that

$$\mathbf{F}_{es}=-G\frac{M_{e}M_{s}}{|\mathbf{r}_{es}|^{2}}\, \hat{\mathbf{r}}_{es}.$$

Here $\mathbf{F}_{es}$ is the force exerted on the satellite due to the earth, and $\mathbf{r}_{es}$ is the radius vector from the earth to the satellite. Therefore, by Newton's Second Law, we have that
$$\mathbf{F}_{es}=M_{s}\mathbf{a}_{s},$$
and hence
$$\mathbf{a}_{s}=-G\frac{M_{e}}{|\mathbf{r}_{es}|^{2}}\, \hat{\mathbf{r}}_{es}.$$
So $\mathbf{a} = -9.8m/sec^2$?

#### Ackbach

##### Indicium Physicus
Staff member
So $\mathbf{a} = -9.8m/sec^2$?
No. $|\mathbf{a}_{g}|=9.8\,\text{m/s}^{2}$ only near the earth's surface. That is, only where the radius vector's length is approximately the radius of the earth. A satellite is going to be much farther away from the earth's center than that. Expect this acceleration to be much lower, in accordance with the inverse square law.

#### MarkFL

Staff member
I believe we would have:

$\displaystyle \mathbf{a}\approx-9.8\left(\frac{r}{r+300} \right)^2\frac{\text{m}}{\text{s}^2}$

where $r$ is the radius of the Earth in km.

#### dwsmith

##### Well-known member
How does one account for the sun by finding the magnitude of the gravitational disturbance caused by the Sun?

#### Ackbach

##### Indicium Physicus
Staff member
How does one account for the sun by finding the magnitude of the gravitational disturbance caused by the Sun?
It would vary greatly depending on the relative alignment of the earth and the sun compared to the satellite. The total gravitational force on the satellite is just the vector sum of the gravitational forces due to the earth and the sun. I think you'll find that the sun's influence is much less than the earth's despite the sun's immensely greater mass.

#### MarkFL

Staff member
The same way you would for the Earth (and find the sum of the forces), except you would have to account for the varying distance of the satellite from the center of the sun since the satellite is orbiting the Earth, and also you would have to account for the varying distance of the Earth from the Sun.

While the size of the region around the Earth in which the Earth's gravity is dominant over that of the sun is very small relative to the solar system, it is large compared to the region in which we place artificial satellites.

#### dwsmith

##### Well-known member
So the equation would be:
$$\mathbf{F}_s = -\frac{GM_sm_s}{r_s^2}\hat{\mathbf{r_s}}$$
where the earth is located at $\theta = \pi$ from the sun and the satellite is at $\beta = \frac{3\pi}{2}$ from the earth.
This force would be positive since it is moving away from earth?
$$\mathbf{a} = \frac{\mathbf{F}_{\text{sunsat}}}{m_{sat}} = -\frac{GM_{\text{sun}}}{r^2_{\text{sunsat}}}\hat{ \mathbf{r}}_{\text{sunsat}}$$
Then $r_s = \sqrt{d_{es}^2 + 300^2}$
Correct?
How do I compare these two(earth and sun on sat) accelerations since I have nothing to compare?

Last edited:

#### topsquark

##### Well-known member
MHB Math Helper
So the equation would be:
$$\mathbf{F}_s = -\frac{GM_sm_s}{r_s^2}\hat{\mathbf{r_s}}$$
where the earth is located at $\theta = \pi$ from the sun and the satellite is at $\beta = \frac{3\pi}{2}$ from the earth.
This force would be positive since it is moving away from earth?
$$\mathbf{a} = \frac{\mathbf{F}_{\text{sunsat}}}{m_{sat}} = -\frac{GM_{\text{sun}}}{r^2_{\text{sunsat}}}\hat{ \mathbf{r}}_{\text{sunsat}}$$
Then $r_s = \sqrt{d_{es}^2 + 300^2}$
Correct?
How do I compare these two(earth and sun on sat) accelerations since I have nothing to compare?
Are you studying this as a class assignment? If so then we need to know of any specifications that your instructor would want. (ie. does your instructor expect you to answer in terms of the angle the Earths's gravitational force on the satellite makes with that of the Sun?)

If not then I would recommend that you pick the satellite at its closest point to the Sun (The satellite is between the Earth and Sun) and at its furthest point (The Earth is between the satellite and the Sun.) I don't know the exact figures, but I'll bet that you can average these two numbers. (They should be about the same amount anyway.)

-Dan

#### dwsmith

##### Well-known member
Are you studying this as a class assignment? If so then we need to know of any specifications that your instructor would want. (ie. does your instructor expect you to answer in terms of the angle the Earths's gravitational force on the satellite makes with that of the Sun?)

If not then I would recommend that you pick the satellite at its closest point to the Sun (The satellite is between the Earth and Sun) and at its furthest point (The Earth is between the satellite and the Sun.) I don't know the exact figures, but I'll bet that you can average these two numbers. (They should be about the same amount anyway.)

-Dan
The picture shows it here:
Code:
\begin{center}
\begin{tikzpicture}
\draw (-4cm,0) -- (4cm,0);
\draw (-4cm,0) circle (1.5cm);
\draw (-4cm,-1.5cm) -- (4cm,-.2cm);
\filldraw[blue] (-4cm,0) circle (.5cm);
\filldraw[orange] (4cm,0) circle (1cm);
\draw[thick] (-4cm,0) node {Earth};
\draw[thick] (4cm,0) node {Sun};
\filldraw[gray] (-4cm,-1.5cm) circle (.1cm) node[below = 1pt] {Satellite};
\end{tikzpicture}
\end{center}

#### topsquark

##### Well-known member
MHB Math Helper
The picture shows it here:
Code:
\begin{center}
\begin{tikzpicture}
\draw (-4cm,0) -- (4cm,0);
\draw (-4cm,0) circle (1.5cm);
\draw (-4cm,-1.5cm) -- (4cm,-.2cm);
\filldraw[blue] (-4cm,0) circle (.5cm);
\filldraw[orange] (4cm,0) circle (1cm);
\draw[thick] (-4cm,0) node {Earth};
\draw[thick] (4cm,0) node {Sun};
\filldraw[gray] (-4cm,-1.5cm) circle (.1cm) node[below = 1pt] {Satellite};
\end{tikzpicture}
\end{center}
(Ahem) Is there any other way you can post that? I don't even know how to run the script.

-Dan

#### dwsmith

##### Well-known member
(Ahem) Is there any other way you can post that? I don't even know how to run the script.

-Dan

#### topsquark

##### Well-known member
MHB Math Helper
Okay so you know how to calculate the gravitational force on the satellite from the Earth.

$$F = \frac{GM_Em}{r^2}$$

The force from the Sun will be the same idea:

$$F = \frac{GM_Sm}{r^2}$$

To get the net result, both forces act in the direction of the objects, toward the Earth or toward the Sun.

One simplification if you want to go down this road: The distance the satellite is from the Earth is practically negligible so feel free to use the distance from the Earth to the Sun.

-Dan

#### dwsmith

##### Well-known member
Okay so you know how to calculate the gravitational force on the satellite from the Earth.

$$F = \frac{GM_Em}{r^2}$$

The force from the Sun will be the same idea:

$$F = \frac{GM_Sm}{r^2}$$

To get the net result, both forces act in the direction of the objects, toward the Earth or toward the Sun.

One simplification if you want to go down this road: The distance the satellite is from the Earth is practically negligible so feel free to use the distance from the Earth to the Sun.

-Dan
I have that but how does one compare the two accelerations since I don't have actual numbers?

#### topsquark

##### Well-known member
MHB Math Helper
I have that but how does one compare the two accelerations since I don't have actual numbers?
I can't figure out a way to do it without putting numbers in. Sorry!

-Dan

#### dwsmith

##### Well-known member
I can't figure out a way to do it without putting numbers in. Sorry!

-Dan
I can never figure this out since G has so many means in my book. Is G a known constant in this equation?

#### MarkFL

Staff member
Yes, $G$ is Newton's universal gravitational constant, which is given by Wikipedia as:

$\displaystyle G\approx6.674\,\times\,10^{-11}\,\frac{\text{N}\cdot\text{m}^2}{\text{kg}^2}$

#### Deveno

##### Well-known member
MHB Math Scholar
Okay so you know how to calculate the gravitational force on the satellite from the Earth.

$$F = \frac{GM_Em}{r^2}$$

The force from the Sun will be the same idea:

$$F = \frac{GM_Sm}{r^2}$$

To get the net result, both forces act in the direction of the objects, toward the Earth or toward the Sun.

One simplification if you want to go down this road: The distance the satellite is from the Earth is practically negligible so feel free to use the distance from the Earth to the Sun.

-Dan
those should be different $r$'s, yes?

#### topsquark

##### Well-known member
MHB Math Helper
those should be different $r$'s, yes?
Yes. The first r will be the distance from the center of the Earth to the satellite and the second r is the distance from the Sun. (No point in worrying about the radius of the Sun, it's very small compared to the distance between the satellite to the Sun.)

-Dan

#### Deveno

##### Well-known member
MHB Math Scholar
well. correct me if i'm wrong, but doesn't this depend on where earth is in its elliptical orbit (aphelion vs. perihelion)?

#### topsquark

##### Well-known member
MHB Math Helper
well. correct me if i'm wrong, but doesn't this depend on where earth is in its elliptical orbit (aphelion vs. perihelion)?
Yes, but given that the distance between the Sun and the satellite is so much larger than the change between aphelion and perihelion it is something that can be ignored if you wish.

-Dan