dr/dt=(-64/5)(G^3/c^5)(m1*m2)((m1+m2)/r^3)
t=r^4*[15*5/(64*16*4)]*[(c^5/G^3)/(m1*m2(m1+m2))]
=> t = (60*10^3)^4*(75/4096)*((3*10^8)^5/((6.673*10^-11)^3)/((1.4*1.989*10^30)^2*(2*1.4*1.989*10^30))
=> t = (6)^4*(75/4096)*((3)^5/((6.673)^3)/((1.4*1.989)^2*(2*1.4*1.989))*10^(16+40+33-60-30)
dr/dt=(-64/5)(G^3/c^5)(m1*m2)((m1+m2)/r^3)
t=r^4*[15*5/(64*16*4)]*[(c^5/G^3)/(m1*m2(m1+m2))]
=> t = (60*10^3)^4*(75/4096)*((3*10^8)^5/((6.673*10^-11)^3)/((1.4*1.989*10^30)^2*(2*1.4*1.989*10^30))
=> t = (6)^4*(75/4096)*((3)^5/((6.673)^3)/((1.4*1.989)^2*(2*1.4*1.989))*10^(16+40+33-60-30)
You can calculate this with Kepler's Third Law. "The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit." This is valid for other orbiting objects; in this case you can replace "planet" with "satellite". Just assume, for simplicity, that the satellite orbits Earth in a circular orbit - in this case, the "semi-major axis" is equal to the distance from Earth's center. For your calculations, remember also that if the radius is doubled, the total distance the satellite travels is also doubled.
The electron and the outermost electrons' orbital radius.
Given a constant velocity v, a radius r, the equation for the period T is given as:T = 2πr/vSo, as radius r increases, so does the period T.
m x (v)squared/F where m is mass, v is the velocity .. this value must be squared. F is the Force
The relationship between those four can be found from using the original centripetal force equation, Fc = (mv2)/r.Since we know v=d/t, we can sub that into the equation to get Fc=(md2)/(rT2), where T is actually the period.Now, we know the distance it travels is in a circular motion, so we can assume the distance it travels is equal to the circumference of that circle. Since we know that equation to be d=2Ï€r, we can sub that into our equation to make Fc=(m[2Ï€r]2)/(rT2). Expand that square brackets to make Fc=(m4Ï€2r2)/(rT2). After cancelling one radius from the top and bottom, you are left with the final equation:Fc=(m4Ï€2r)/T2, where m = the mass of the revolving object, r = radius of the curvature, and T = rotation period of the revolving mass.
it is the distance between what an object is orbiting around and the object itself in any given point
Yes. T = (2pi / sqroot of GM) multiplied by the radius^3/2. A planets mass DOES NOT affect its orbital period. A planets radius DOES affect its orbital period.
It doesn't orbit earth faster. The ISS is in a lower orbit with a period of 91 minutes compared to the Hubble's orbital period of 96-97 minutes. Orbital periods generally increase with orbit radius and speed in the orbit decreases with increasing orbit radius.
The point when an orbiting object is closest to the central object is called periapsis. For the specific case of planets around the Sun, the term perihelion is used as well.
19.2 rE
pluto
Of Jupiter or the Sun? Technically it is orbiting both. It orbits the Sun at a radius of (on average) 5.204 AU or about 780 million kilometres. Titan is a moon of Saturn (not Jupiter) and orbits Saturn at approximately 1.2 million kilometers radius. Saturn in turn orbits the sun at about 10 A.U. or 1.5 billion kilometers.
yeet
The equation for circumference is C=2(pi)(r) where C is the circumference and r is the radius. You can find the radius by plugging in the circumference into this equation and solving the equation.
That's the approximate radius of a neutron star, a.k.a. a pulsar.
Assuming a circular orbit for simplicity, the magnitude of the angular momentum is rmv - that is, the radius of the orbit times the mass times the velocity. I'll leave the details of the calculations to you; basically you have to look up:Earth's, or the Moon's, orbital radius (the distance from Sun to Earth vs. the distance from Earth to the Moon);The mass of the orbiting object;Its velocity in orbit.Then you must divide one by the other, since I assume it's the ratio you are interested in.
You can calculate this with Kepler's Third Law. "The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit." This is valid for other orbiting objects; in this case you can replace "planet" with "satellite". Just assume, for simplicity, that the satellite orbits Earth in a circular orbit - in this case, the "semi-major axis" is equal to the distance from Earth's center. For your calculations, remember also that if the radius is doubled, the total distance the satellite travels is also doubled.