Physics Help Forum Gravitational and Electric fields

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 Jun 16th 2019, 04:36 PM #1 Junior Member   Join Date: Jun 2019 Posts: 3 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? Last edited by Adam97; Jun 16th 2019 at 05:21 PM.
Jun 16th 2019, 05:34 PM   #2

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 Originally Posted by Adam97 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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Jun 16th 2019, 05:40 PM   #3
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 Originally Posted by topsquark $\displaystyle F = \dfrac{G M_s ~ M_m}{r^2}$ 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.

Jun 16th 2019, 06:20 PM   #4

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 Originally Posted by Adam97 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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Jun 16th 2019, 06:25 PM   #5
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 Originally Posted by topsquark 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.

Jun 16th 2019, 07:33 PM   #6

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 Originally Posted by Adam97 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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 Jun 18th 2019, 05:38 PM #7 Senior Member   Join Date: Mar 2019 Location: cosmos Posts: 558 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...
Jun 18th 2019, 07:36 PM   #8

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 Originally Posted by neila9876 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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