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cdangel · Oct 28, 2009, 06:04 PM

All right guys, here's a really hard one I cannot get.

A newly-discovered planet has only one moon. Their center-to-center distance is 790 000 000 metres. The planet's mass is 90 times the mass of its moon. How far from the center of the planet would a space shuttle have to be so it would experience no net force?

This is all the info I got.

Any help?
Possible equations?

Nhatkiem · Oct 28, 2009, 06:28 PM

Originally Posted by cdangel

Alright guys, here's a really hard one I cannot get.

A newly-discovered planet has only one moon. Their center-to-center distance is 790 000 000 metres. The planet's mass is 90 times the mass of its moon. How far from the center of the planet would a space shuttle have to be so it would experience no net force?

This is all the info I got.

Any help?
Possible equations?

Gravitational force between two objects is explained as

$F_g=G\frac{m_1m_2}{r^2}$

where m is the mass of the two objects, G is the gravitational constant (6.673*10^-11) and r is center to center distance between the two objects.

We then want the shuttle to be in a position where the force exerted by the moon and the planet are equal! (thus net force is zero).

let the following be true:
mm = mass of the moon
mp = mass of the planet
ms = mass of the ship
fm = force between moon and ship
fp = force between planet and ship

as stated above, the forces must be equal, therefore:

$F_m=F_p$

$G\frac{m_mm_s}{r_1^2}=G\frac{m_pm_s}{r_2^2}$

Now if we think about r here, if the ship is between who objects that are only 100m away, we say that if it is 10m from object a, then it must be 90m from object be. Or if the distance equals x from a, then the distance must be 100-x from b. so here we can write that the distance the ship is from the moon be r, and the distance from the planet be 7.9*10^8-r. Therefore we have:

$G\frac{m_mm_s}{r^2}=G\frac{m_pm_s}{(7.9*10^8-r)^2}$
Lets start cancelling!
$\frac{m_m}{r^2}=\frac{m_p}{(7.9*10^8-r)^2}$

But wait! Were also given that the mass of the planet is 90 times the moon, so we can say:

$m_p=90*m_m$
we then have:
$\frac{m_m}{r^2}=\frac{90*m_m}{(7.9*10^8-r)^2}$
Factor the bottom out and bit more cancellation with mm
$\frac{1}{r^2}=\frac{90}{(7.9*10^8)^2-2r(7.9*10^8)+r^2)}$
Some algebra work:
$(7.9*10^8)^2-2r(7.9*10^8)+r^2)=90r^2$
Combine like terms!
$0=89r^2+2r(7.9*10^8)-(7.9*10^8)^2$

This becomes just a quadratic equation :cool:, solve for r

Nhatkiem · Oct 28, 2009, 06:32 PM

I forgot to say, If you know calculus, then just take the derivative with respect to r, and solve it as an optimization problem.

cdangel · Oct 29, 2009, 07:00 PM

Originally Posted by Nhatkiem

Gravitational force between two objects is explained as

$F_g=G\frac{m_1m_2}{r^2}$

where m is the mass of the two objects, G is the gravitational constant (6.673*10^-11) and r is center to center distance between the two objects.

We then want the shuttle to be in a position where the force exerted by the moon and the planet are equal! (thus net force is zero).

let the following be true:
mm = mass of the moon
mp = mass of the planet
ms = mass of the ship
fm = force between moon and ship
fp = force between planet and ship

as stated above, the forces must be equal, therefore:

$F_m=F_p$

$G\frac{m_mm_s}{r_1^2}=G\frac{m_pm_s}{r_2^2}$

Now if we think about r here, if the ship is between who objects that are only 100m away, we say that if it is 10m from object a, then it must be 90m from object be. Or if the distance equals x from a, then the distance must be 100-x from b. so here we can write that the distance the ship is from the moon be r, and the distance from the planet be 7.9*10^8-r. Therefore we have:

$G\frac{m_mm_s}{r^2}=G\frac{m_pm_s}{(7.9*10^8-r)^2}$
Lets start cancelling!
$\frac{m_m}{r^2}=\frac{m_p}{(7.9*10^8-r)^2}$

But wait! were also given that the mass of the planet is 90 times the moon, so we can say:

$m_p=90*m_m$
we then have:
$\frac{m_m}{r^2}=\frac{90*m_m}{(7.9*10^8-r)^2}$
Factor the bottom out and bit more cancellation with mm
$\frac{1}{r^2}=\frac{90}{(7.9*10^8)^2-2r(7.9*10^8)+r^2)}$
Some algebra work:
$(7.9*10^8)^2-2r(7.9*10^8)+r^2)=90r^2$
Combine like terms!
$0=89r^2+2r(7.9*10^8)-(7.9*10^8)^2$

This becomes just a quadratic equation :cool:, solve for r

Thanks a bunch for that.
NOW I'm going to seem like a total noob, but when i solve for r using the quadratic Formula, both the answers end up being negative, which I find to make no sense.

A little bit more help here please?

Nhatkiem · Oct 29, 2009, 07:56 PM

Originally Posted by cdangel

Thanks a bunch for that.
NOW I'm going to seem like a total noob, but when i solve for r using the quadratic Formula, both the answers end up being negative, which I find to make no sense.

I'm not sure what your doing wrong because I am getting a negative and a positive answer. Make sure you are typing in everything correctly :confused:

Unknown008 · Oct 29, 2009, 11:51 PM

I'll point out that the equation is in the form : ax^2 + bx - c

No matter how you do, you must have a positive and a negative factor, because the factors of - c are a positive number and a negative number. Post what you did, perhaps we can help you spot your mistake, or you'll find it yourself while copying it. :)

Nhatkiem · Oct 29, 2009, 11:58 PM

Originally Posted by Unknown008

I'll point out that the equation is in the form : ax^2 + bx - c

No matter how you do, you must have a positive and a negative factor, because the factors of - c are a positive number and a negative number. Post what you did, perhaps we can help you spot your mistake, or you'll find it yourself while copying it. :)

Lol, the original problem kept biting me in the butt because my radius kept canceling each other out haha. But you have a point, the answer will have to be 1 negative and 1 positive value, two factors of the same sign would yield positive c.

cdangel · Oct 30, 2009, 12:12 PM

so exactly what answers did you guys get?
cause I ended up with about 7.5 x 10^7 as the distance from the moon

Unknown008 · Oct 30, 2009, 12:23 PM

Well, I got that too. If I'm taking Nhatkiem's work as being good (I trust him enough, so I didn't check, I don't have the patience :o) then, yes, it's good.

Nhatkiem · Oct 30, 2009, 03:36 PM

Originally Posted by Unknown008

Well, I got that too. If I'm taking Nhatkiem's work as being good (I trust him enough, so I didn't check, I don't have the patience :o) then, yes, it's good.

Of course the answer is right :p

Anyway that was also the answer I got.

Ideally you would imagine the object would have to be closer to the moon to feel no forces since the moon is smaller. So the distance you got would have to be the distance with respect to the moon since its only about 10% of the total distance r.