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Old Jun 16th 2019, 04:36 PM   #1
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Angry Gravitational and Electric fields

Question.

The sun exerts a gravitational force of 1.64 x 10^21 N on Mars. Calculate the mass of Mars. [ Mass of the Sun = 1.989 x 10^30 kg. Mean seperation of Mars from the Sun = 2.279 x 10^11 m.]

Can anyone help me with this?

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Old Jun 16th 2019, 05:34 PM   #2
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Originally Posted by Adam97 View Post
Question.

The sun exerts a gravitational force of 1.64 x 10^21 N on Mars. Calculate the mass of Mars. [ Mass of the Sun = 1.989 x 10^30 kg. Mean seperation of Mars from the Sun = 2.279 x 10^11 m.]

Can anyone help me with this?
$\displaystyle F = \dfrac{G M_s ~ M_m}{r^2}$

So, G is G, you are given the mass of the Sun ( $\displaystyle M_s$ ), and the distance between them r. So what is the mass of Mars, $\displaystyle M_m$?

What does this have to do with electric fields?

-Dan
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Old Jun 16th 2019, 05:40 PM   #3
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Originally Posted by topsquark View Post
$\displaystyle F = \dfrac{G M_s ~ M_m}{r^2}$

So, G is G, you are given the mass of the Sun ( $\displaystyle M_s$ ), and the distance between them r. So what is the mass of Mars, $\displaystyle M_m$?

What does this have to do with electric fields?

-Dan
I'm not sure myself. The unit im doing currently in physics is Gravitational and Electric Fields. This is one of the questions we have in the assignment.
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Old Jun 16th 2019, 06:20 PM   #4
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Originally Posted by Adam97 View Post
I'm not sure myself. The unit im doing currently in physics is Gravitational and Electric Fields. This is one of the questions we have in the assignment.
Were you able to get the mass of Mars?

-Dan
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Old Jun 16th 2019, 06:25 PM   #5
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Originally Posted by topsquark View Post
Were you able to get the mass of Mars?

-Dan
No not yet. Thats what its asking me to do but can not figure out how.
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Old Jun 16th 2019, 07:33 PM   #6
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Originally Posted by Adam97 View Post
No not yet. Thats what its asking me to do but can not figure out how.
Get $\displaystyle M_m$ by itself:
$\displaystyle F = \dfrac{G M_s ~ M_m}{r^2}$

$\displaystyle r^2 F = G M_s M_m$

$\displaystyle \dfrac{r^2 F}{G M_s} = M_m$

Can you finish?

-Dan
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Old Jun 18th 2019, 05:38 PM   #7
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Talking

Putting gravity and electricity in one unit is not a bad idea. People always want ot seek a way to unite them (or say to find a way to relate them together). We talked about this respect for a short while in Woody's thread "black hole image" in the lounge columne too...
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Old Jun 18th 2019, 07:36 PM   #8
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Originally Posted by neila9876 View Post
Putting gravity and electricity in one unit is not a bad idea. People always want ot seek a way to unite them (or say to find a way to relate them together). We talked about this respect for a short while in Woody's thread "black hole image" in the lounge columne too...
BUT this has nothing to do with the conversion at hand. I think it is good that you want to refer the OP to another thread they might be interested in, but you should do that over the PM system.

-Dan
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