Physics help....?

Question

I have a question I have been working on for a while about a the earth and the moon. I am supposed to find the total acceleration caused by the earth and moon at a certain point. I am given the mass of earth (5.98x10^24 kg) and the mass of the moon (7.36x10^22). The earth, moon and the point make up a triangle. The distance from earth to the moon is 3.84x10^8 m and the moon to the point is 1.48x10^8 m and the distance from the point to earth is 3.54x10^8 m. Also I have to find the magnitude of acceleration at this point. The last part is What is the total gravitational force on a spacecraft and the point? The mass of the spacecraft is 1200kg . What is the magnitude of the total gravitational force on the spacecraft.

If anyone can even steer me in the right direction it would help a lot...Thanks.

Answer 1

The simple answer: the net force at the point is equal to the vector sums of the gravitational forces exerted by the Earth and the moon.

The long, tricky answer: trig and careful problem solving.

STEP 1:

The first step to solving this problem is to state your assumptions. From the problem statement, we can make a few simplifying assumptions to make this analysis much easier:

A1. No additional forces (magnetic, electrostatic, etc) are acting at the point.

A2. The positions of the Earth, moon and point are fixed with respect to each other (no orbiting)

A3. The point itself is massless, and the mass of the spaceship is much less than the mass of the Earth


STEP 2:

Now let's define our variables:

P = Name of arbitrary point
RE = Radius of the Earth (6.373x10^6 m)
RM = Radius of the moon (3.475x10^6 m)
a = Distance from the Earth to the arbitrary point P
b = Distace from the moon to the arbitrary point P
c = Distance from the Earth to the moon
alpha = Angle between "b" and the x-axis
beta = Angle between "a" and the x-axis
gamma = Angle between "a" and "b"
ME = Mass of the Earth
MM = Mass of the moon
MS = Mass of the spacecraft
G = Gravitational Constant (6.6742x10^-11)


STEP 3:

The next step is to draw it out correctly and set up your coordinate system. Since part of the problem deals with a spacecraft, let's make this easy by placing Earth at the origin. Then we'll place the moon at some distance "c" along the +x-axis. Finally, we'll place the arbitrary point at some point (x, y) where x is between 0 and c.


STEP 4:

We can now solve for the angles alpha, beta and gamma by using the law of cosines. This law states that:

a² = b² + c² - 2bc*cos(alpha)
b² = a² + c² - 2ac*cos(beta)
c² = a² + b² - 2ab*cos(gamma)

Solving for the angles:

alpha = arccos((a² - b² - c²)/(-2bc))
beta = arccos((b² - a² - c²)/(-2ac))
gamma = arccos((c² - a² - b²)/(-2ab))


STEP 5:

Now that we've layed out our assumptions (step 1), made a diagram of the system (step 2), labeled all of the terms (step 3) and defined their relationships to one another (step 4), we will now apply this information to the laws of gravity to obtain the final answers.

According to Newton's Law of Gravity, the gravitational force exerted on one body by another is given by the equation:

F = Gm1m2/r²

where:

G = Gravitational Constant
m1, m2 are the masses of each body
r = the distance between the center of mass of each body (as opposed to the surface of each body)

The acceleration of mass 2 due to mass 1 is thus:

a2 = Gm1/r²

Note that the direction of the force and acceleration is in a direct line-of-sight between the two bodies.


STEP 6:

Finally, we can apply all of this information to solving the problem. The magnitude of the acceleration at P due to the Earth is:

a1 = G(ME)/(a + RE)²

This is the magnitude of the acceleration vector. It is directed towards the Earth. Thus, by using more trig and the coordinate system we initially defined, we find that the x and y values of this vector are:

a1x = G(ME)/(a + RE)² * cos(beta)
a1y = -G(ME)/(a + RE)² * sin(beta) (notice the negative sign)

Similarly, the acceleration vector on the point caused by the moon is:

a2 = G(MM)/(b + RM)²

The components of this vector are:

a2x = G(MM)/(b + R)² * cos(alpha)
a2y = -G(MM)/(b + RM)² * sin(alpha) (notice the negative sign)

Finally, the net acceleration of the point can be found by observing that the y-components of each acceleration are in the same direction (so they add) and the x-components are in opposite directions (so they subtract). Thus, the net acceleration of the point is given by:

ay = a1y + a1y
= -G(ME)/(a+RE)²*sin(beta) - G(MM)/(b+RM)²*sin(alpha)

ax = a1x - a2x
=G(ME)/(a+RE)² *cos(beta) - G(MM)/(b + R)² * cos(alpha)

Where the magnitude of the total acceleration is found using the Pythagorean Theorem:

a = sqrt(ax² + ay²)

where the direction of travel is simply:

theta = arctan(ay/ax)

To find the gravitational force on the spaceship, you have to multiply the mass of the ship by the net acceleration at that point:

F = MS*a


CONCLUSION:

The solution to the problem has now been placed entirely in terms of known variables. From this point, it is simply a matter of "plug and chug" mathematics. The numbers I get are, roughly:

alpha = 67.2°
beta = 22.8°
gamma = 90°

ay = -.00139 m/s²

ax = .00275 m/s² (positive value indicates motion towards the Earth)

theta = arctan(ay/ax) = -26.82°

a = .003081 m/s² @ -26.82° below the horizon

F = 3.7 N (tending towards the Earth)

Answer 2

There are two ways to approach this and they are pretty much equal in the amount of work you will have to do.

Start by drawing a picture, a triangle with the three points (earth, moon and object point) identified. You know the lengths of all three sides so you can uniquely construct a triangle.

For the first way, find the acceleration vector attracting the object point to each planetary body. Then, add the vectors to find the total acceleration.

For the second way, find the center of gravity between the moon and the earth. The gravitational acceleration is directed at that center of gravity and the effective mass is the sum of earth and moon mass.

The last part of your problem is trivial once you have found the acceleration.

Answer 3

try using this formula F=Gm1m2/r^2 where m1 and m2 are the masses and the G is the gravitational constant. This shows the gravitational pull they have. Also to be completely accurate you need the radius of the earth and moon because to measure gravitational pull, it must be done from the center of the objects. It seems like they over looked that though. Because they are at angles you may need to use sin or cos to get the magnitudes.

Answer 4

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Answer 5

Hi. You might look at Lagrangian point : http://www.answers.com/topic/lagrangian-point
as a start.