The distance between the earth and the moon is 3.85x10^8m.Find the time it took for the first person who walk?
Radio waves run at the speed of light, which is 3.0x10^8 m/s. So, it takes just over 1 second for radio to go from the moon to earch. (1.28 to be more exact.) I got that by dividing 3.85 / 3.0 (the 10^8's cancel). So 5.6x10^10 / 3.0x10^8 is 186.67s. Or, just over 3 minutes.
Do you think the earth will have a second moon someday?
Absolutely. At some point in the not so far off future we will capture small asteroids, like 2007 TU24, as they pass by earth. We will harvest them for raw materials to build equipment for spacetravels and off world operations. As these asteroids will orbit earth just like the moon they will infact be moons. Many planets in the solarsystem have captured asteroids for moons. Mars has two. Ours will just be captured by us humans.
The masses of Earth and Moon are 6*10^24 Kg and 7.4*10^22 Kg . The distance between them is 3.84 * 10^5 Km. Claculate it's gravitational force of attraction between the two?
Please, do yourself a favor and employ your little grey cells[1] now and then. Thinking is not only for nerds. Whether you are a jock, a cheerleader, an emo (is this still a thing?) artsy type, or whatever, trust me, thinking does not hurt, and can be fun.F = 6*10^24 x 7.4*10ˇ22 x 6.67*10^-11 / (3.84*10^8)^2Pull out your calculator, mind the precedence of operations, and note conversion from kilometers to meters.Also note the value of gravitational constant G:(BTW, I remember with fondness my fumbling with a variation on Cavendish experiment in attempts to measure the value of G. I barely got the order of magnitude right.)Footnotes[1] Hercule Poirot - Wikipedia
The mass of the Moon is 7.36 × 10[math]^{22}[/math] kg and its distance to the Earth is 3.84 × 10[math]^{8}[/math] m. What is the gravitational force of the Moon on the Earth?
Using newtons law of gravitation, F=GMm/r^2 we could find the gravitational attraction between the moon and Earth. G is the constant 6.67*10^-11 Nm^2/kg^2.F=(6.67*10^-11)(5.97*10^24)(7.36*10^22)/(3.84*10^8)^2F=1.9875384*10^20 NewtonsRemember to use the correct units.
A spacecraft is on a journey to the moon. At what point, as measured from the center of the earth, does the gr?
The gravitational force is proportional to the mass of the body and inversely proportional to the square of the distance. Assume that the distance between earth and moon is "1" and the spacecraft is "x" distance from earth. Then: F(e) ~ M(e)/x^2 F(m) ~ M(m)/(1-x)^2 Therefore at the point they are the same: M(e)/x^2 = M(m)/(1-x)^2 M(e)(1-x)^2 = M(m)x^2 M(e)(x^2 -2x +2) = M(m)x^2 (x^2 -2x +2) = (1/81.4)x^2 .9877x^2 -2x +2 = 0 This has two roots, only one of which is between 0 and 1. x~.9002 So the forces are equal when the craft is 90% of the way to the moon. .90 * 3.85x10^8m 3.47x10^8m from earth
What is the angular momentum of the moon around the earth?
The Moon’s angular momentum with respect to the Earth comes from two sources: first, its orbit around the Earth; and second, from its spin around its own axis.Let us first suppose that:The Moon is a uniformly dense sphere (it isn't) of radius a 1.7375e3 km,The Moon’s mass M is 7.3459e22 kg,The Moon’s orbital period T is 27.322 days, or 2.3606e6 s,The Moon’s rotational period T is 27.322 days, or 2.3606e6 s, andThe Moon’s orbit is circular (it isn't), and is 385,000 km in radius r.The moon’s angular velocity (both for its orbit and its rotation, since its tidally locked in a 1:1 orbit) is:ω = 2π / T = 2π / 2.3606e6 s = 2.66169e-6 rad/sThe Moon’s rotational moment of inertia is:I = 2/5 M a^2 = 0.4(7.3459e22 kg)(1.7375e6 m)^2OrI = 8.87063e34 kg m^2So the rotational angular momentum L is:L = Iω = (8.87063e34 kg m^2)(2.66169e-6 rad/s)OrL = 2.361087e29 kg m^2/sNow compare that to the Moon’s orbital moment of inertia:I = Mr^2 = (7.3459e22 kg)(3.85e8 m)^2OrI = 1.0888e40 kg m^2And its correspondingly larger orbital angular momentum:L = Iω = (1.0888e40 kg m^2)(2.66169e-6 rad/s)OrL = 2.89805e34 kg m^2/s, about 122,741 times larger!
A spacecraft is on a journey to the moon. At what point, as measured from the center of the earth, does the gr?
Let d = 3.85 * 10^8 m; Me = mass of the earth Mm = mass of the moon = Me/81.4 m = mass of the satellite G = gravitational constant Fe = gravitational force due to the earth Fm = gravitational froce due to the moon Then Fe = GMem/r² where r is the distance from the centre of the earth and Fm = GMm*m/(d-r)² or Fm = GMem/(81.4*(d-r)²) where we used Mm = Me/81.4 Where gravity from the earth and moon balance eath other, we have Fe = Fm or GMem/r² = GMem/(81.4*(d-r)²) Cancel G, Me and m from both sides to get 1/r² = 1/(81.4*(d-r)²) d is known so solve for r r² = 81.4*(d-r)² = 81.4(d² - 2dr + r² ) 80.4r² - 162.8dr + 81.4d² = 0 Use the quadratic formula to solve this equation. You get two results: r = 3.47 * 10^8 m and re = 4.33 * 10^8m Note that distance re is on the far side of the moon where the forces are equal but not balanced. We are looking for the balanced solution between the earth and the moon. So r = 3.47 * 10^8 m is the solution. Hope this helps.
How long does it take for a radio signal from Earth to reach the moon?
The Moon is 384,400 kms from Earth. The speed of light is 299,792 kms per second.Therefore 384,400÷299,792=1.282 sec for a radio signal to travel to the moon.So when having a conversation with a person on the moon the delay will be doubled to 2.564 sec.
Please help me with a physics problem I dont understand rotational motion and angular displaement?
this question does not deal with motion, but rather with angle draw diagrams to show that you can determine the angular size of each object; you will want to see if either the dime or pea covers a larger angle than the moon in each case, tan(theta) = size of object/distance from your eye for the pea: tan(theta) = 0.5cm/71cm = 0.007 for the dime, tan(theta) = 1.8cm/71 cm = 0.0254 for the moon, tan(theta) = 3.48x10^6m / 3.85x10^10 m = 9.0x10^-5 so both the pea and dime will cover the moon
Is it true that zero gravity field exists somewhere between earth and moon gravitational field?
You can use Newton's universal law of gravitation to find out the answer.Formula:F = GMm/r^2Data:Mass of earth(Me) = 5.97E24 kgMass of moon(Mm) = 7.32E22 kgAverage distance between earth and moon(r) = 384400 kmSolution:Let the point where the net force equals zero be 'x' km from earth, then,G(Me)m/x^2 = G(Mm)m/(r-x)^2=>Me/x^2 = Mm/(r-x)^2=>(r-x)^2/x^2 = Mm/MeSolving and substituting the data, we getx = 346306 kmHence, at 346306 km from earth, net force on a body due to earth's and moon's gravitational field equals 0