Home work help.. What is the gravitational force between the earth and the moon?...?
1. "what is the gravitational force between the earth and the moon?". okay i'll help on these, but i won't give you the answers. :) for calculating the gravitational attraction (or force) between two objects, we use our good friend Sir Isaac Newton. his formula for this problem is: F = G * m(1) * (m2) / r^2 F is the answer. it is the gravitational force between the two objects your comparing ( m(1) and m(2) ) G is the gravitational constant, 6.6726 x 10^-11 m(1) is the mass of one of the objects you comparing. in your case, it could be defined as either the moon or the earth. take your pick! but on m(2), you have to use the other object. it wouldn't make sense if m(1) and m(2) was earth. make sure this is in kg! m(2) is the mass of the second object. again, make sure its in kg and not pounds or tons! r is the distance between the two objects. this value should be in meters. so lets define the variables. F = ? G = 6.6726 x 10^-11 m(1) = mass of earth (5.9742 × 10^24 ) m(2) = mass of moon (7.36 × 10^22) r = distance between moon and earth (402,336,000 meters) you might want to get a calculator for this. so lets rewrite the formula, but this time, including the values. F = .000000000066726 x 5974200000000000000000000 x 73600000000000000000000 / 402336000 ^2 the just calculate! and thats your answer. the force should be labeled with "N" for newtons, the measurment of force. and gravity is a force! 2. for your second question, we use Kepler's third law to calculate the mean (average) orbital period. his formula for this is: T= √(4π^2 (r^2) / GM T is the time (?) r is the mean satelite orbital radius. G is the gravitational constant (6.6726 x 10^-11) M is the mass of the planet the moon is orbiting around. so lets define our variables with values. T = ? r = 402,336,000 (the average distance of the moon from the earth) M = 5974200000000000000000000 kg so plug in all the values. T= √4(3.14 ^2) (402336000 ^2) / (.000000000066726* 5974200000000000000000000 ) get a calculator for that one too! hope this helps!
How far from Earth must a space probe be along a line toward the Sun so that the Sun gravitational pull on the probe balances Earth's pull?
Pull of gravity: F = G M m / d^2 This is the general formula, where G is a constant coefficient (it is there to make the units work out), M is the mass of one object (usually the more massive one), m is the mass of the smaller object (in this problem, it will be the probe), and d is the distance. Usual units are kg (for mass) and metres (for distance). Instead of M, let's use E for the mass of Earth, and S for the mass of the Sun Let's use x for the distance Earth-probe (what you are looking for) and (150,000,000,000 - x) for the distance Sun-probe (based on the approximation of 150 million km for the Earth-Sun distance) The force of gravity caused by Earth F = G E m / x^2 The force of gravity caused by Sun F = G S m / (1.5*10^11 - x)^2 You want one to cancel the other one out (you want the absolute values of F to be equal) G E m / x^2 = G S m / (1.5*10^11 - x)^2 G and m cancel out (the have the same value on both sides) move S to the left, move x^2 to the right (E/S) = x^2 / (1.5*10^11 - x)^2 The ratio of the Sun's mass to Earth's mass is 332946. (1 / 332,946) = x^2 / (1.5*10^11 - x)^2 take square root both sides 1/577 = x/(1.5*10^11 - x) cross-multiply 1.5*10^11 - x = 577 x 1.5*10^11 = 578x
Is the Sun's gravitational pull on Jupiter stronger than on Earth (due to its mass)?
The gravitational force between two objects is defined by the equation: F = G*(m1*m2)/r² where m1 = mass of object 1 m2 = mass of object 2 r = radius between the centers of mass of the two objects G = gravitational constant = 6.672x10^-11 m³/kg-sec² The value of r for the Earth-Sun system = 1.496 x 10^11 m = Res The value of r for the Jupiter-Sun system = 7.783 x 10^11 m = Rjs Mass of the Earth = 5.9742 x 10^24 kg = Me Mass of Jupiter = 1.899 x 10^27 kg = Mj Mass of the Sun = 1.9891 x 10^30 kg = Ms So the ratio of Jupiter-Sun/Earth-Sun = Ratio = [G*(Mj*Ms)/Rjs²]/[G* (Me*Ms)/Res²] Simplifying, we find the G cancels out, as well as Ms, and then the equation simplifies to: Ratio = (Mj*Res²)/(Me*Rjs²) Substituting in our values from above, we get: Ratio = [1.899 x 10^27 kg * (1.496 x 10^11 m)²]/ [5.9742 x 10^24 kg * (7.783 x 10^11 m)²] Ratio = 11.74 That uses actual numbers from the reference. If we use the mass ratio given in the problem, let's see what turns up: Ratio = 318 * (1.496 x 10^11 m)²/(7.783 x 10^11 m)² = 11.75 As noted above, Jupiter is ~5 times farther from the Sun -- a more precise figure from the reference is 5.20283, so let's plug that in: Ratio = 318/(5.20283)² = 11.75 Finally, the actual ratio of the masses is 317.83, so plugging that in to the equation immediately above gives: Ratio = 317.83/(5.20283)² = 11.74 So, the gravitational force between Jupiter and the Sun is nearly twelve times that of the force between the Earth and Sun.
Which force is greater over the distance between the earth and the sun?
Both forces decrease at the same rate, inversely with the square of the distance. The weakest of the four fundamental forces of nature is gravity. If the Earth were composed of all negatively charge particles and the Sun all positively charged particles, to orbit the Sun, the Earth would have to be at a distance of over 15 billion light years from the Sun. In answer to your question, the electromagnetic force is greater at any distance. Since the charges within the Sun balance out and also those within the Earth, they cancel and gravity is the force remaining and that keeps Earth orbiting. .
Is there any force acting between planets and the sun, when the planets are moving around the sun?
Yes, there are two forces that act when planets revolve around the sun. These are namely gravitational force and centrifugal force. I will try to explain the nature of these forces. The gravitational force is exerted by the sun on the planets which occurs because of the mass of the sun. The centrifugal force is exerted by the rotating motion of the planets which acts in a tangential direction to its elliptical orbit around the sun. The situation with these two forces is that they delicately balance each other. Consider this, if only the gravitational pull of the sun would be present in this system then the planets would start to move towards the sun and ultimately collapse in it. On the other hand if only the centrifugal force were present then because of its tangential nature the planets would slowly move out of their orbit and float away in vacuum. So the reason why the planets move around the way they do is that the gravitational pull of the sun is just balanced by the planets centrifugal force. The path of the planet can actually be calculated and mapped using the balancing equation between the two forces. This is how the primary forces in our solar system work.
If the distance between earth and sun were doubled does the change in effect of gravitational force occur instantly?
If you take the Earth to double the distance from where it is situated now... the gravitational attraction would break into 1/4th of what it was.The formula of gravitational force is F = (G × M1 × M2) / R^2 where, F is the force of gravity, M1 is the mass of the 1st body, M2 is the mass of the other body and R is the distance between the two bodies.If you want to keep the Earth in that new orbit... you have to increase the speed of Earth's revolution considerably .Hope it helps :)
If the Sun's gravity is constantly pulling planets toward it, why hasn't the Earth been pulled into the Sun?
There's a few questions here that has been avoided very cutely by the "theories". My definition for a theory is "the flight of fancy of a demented mind". People who develops theories usually can't see an ice-cream for what is. There is only truth and un-truth, reality and non-reality/childish dreams.Firstly: When the earth was caught by the sun's gravity, why did it happen at all? How in this and all other universes could this happen? What is the chance of a planet(with life and all), or no, 9 planets cruising past this sun, in this galaxy(out of a trillion galaxies and most probably the same amount of different universes(as you all know by now, the black hole theory is gone, everybody that is anybody are now twigging the maths to fit the logic that a black hole is actually a rift or wormhole, to another universes or dimensions?) and getting caught in the sun's gravity force. My definition for the phenomena we call gravity is the following "an inherent pulling force in atoms which increases as atoms conglomerates". With all the examples given, everybody seems to forget that the earth also has a gravitational force? So, to correct the "string" example a bit, let's mount a powerful u-magnet on a centre pivot, take another magnet about a 1000 times weaker and connect it to the end of the string, just where you can feel the smaller magnet start to tug on your hand towards the bigger magnet. Now spin the whole shebang and see what happens! Now you have taken into account that both bodies exerts a pull on each other.Of course, these facts now pull apart all of the theorists dreamy theories. So, if a planet is trapped in a sun's atomic mass attraction force, it has only one way to go, to toast and with a helluva speed!The fact of the matter is that this whole solar system, together with the earth's built in magnetic field generator to stop the sun from incinerating us(Goldilocks is just a children's fairy tale, and will always be!), and even more, using the same generator to create a constant spin which does not vary even with massive gravitational pulls from the sun and the other planets and the moon, can not exist! The whole thing is artificial and some external force is balancing the whole act.
Why don't we feel the Sun's gravity pull?
The Earth's gravitation force is powerful enough because the equation for gravity is squared. That squared factor is the difference. Gravity = Mass times distance squared. If you double the distance from the object, you lessen it's gravitation force upon you by a factor of 4. If you triple your distance from an object, you lessen it's gravitation force upon you by a factor of 9. If you increase the distance by a factor of 10, the gravitation force is lessened by a factor of 100. Now think of this when you consider your distance from the Sun compared to your distance from Earth. The Sun is not 93 million times larger than the earth, and you commonly not more than 93 million microns from Earth, never mind 93 million miles. So, determine the number of microns in 93 million miles and then square that number. Then create a fraction with the digit 1 about that number in the numerator, while that number is in the denominator. For example (1/999,999,999,999,999,999,999,999). This is not the actual answer, but it is the comparable force of the graviational pull of the Sun on you compared to the Earth. That is not going to lift a hair off your head, or pull the free floating atmosphere away from planet Earth. The Lagrange points ( the points in outer space when the gravitation pull of the Sun equals the graviational pull of the Earth are very close to Earth. Study the Lagrange points for more information. You do not have to travel very far from Earth to find the points in space where the graviational force of the Sun is equal to the graviational force of the Earth. Then you say, why doesn't the Sun just pull the Earth into it. The Sun does not pull the Earth into it, because the Earth is in motion around the Sun. If the Earth were to stop rotating around the Sun, the Earth would be pulled into the Sun. Just as if the Space shuttle stopped traveling 18,000 miles per hour around the Earth, it would be pulled back into the earth. If you do not believe me, study the re-entry speed of SpaceShip One recently. This is all basic Newtonian gravity. Any good science teacher would know this.