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)
As the orbital radius increases in a planetary system, the period of the orbiting object also increases. This means that the time it takes for the object to complete one full orbit around its central body becomes longer as the distance between them grows.
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.
If you decrease a planet's orbital radius, its surface temperature will increase.
Circular orbital motion refers to the motion of an object around a central point in a circular path under the influence of a centripetal force. This type of motion is commonly observed in celestial bodies like planets orbiting around stars. The speed and radius of the orbit determine the behavior of the object in circular orbital motion.
The formula to find the orbital speed v for a satellite in a circular orbit of radius r is v (G M / r), where G is the gravitational constant, M is the mass of the central body, and r is the radius of the orbit.
it is the distance between what an object is orbiting around and the object itself in any given point
As the orbital radius increases in a planetary system, the period of the orbiting object also increases. This means that the time it takes for the object to complete one full orbit around its central body becomes longer as the distance between them grows.
Using Newton's law of universal gravitation and the fact that the gravitational force is equal to the centripetal force, we can set up an equation to find the orbital radius of Los. By equating the gravitational forces of both Los and Ratipuj, we can find the orbital radius of Los to be approximately 1.49 x 1010 km.
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.
If you decrease a planet's orbital radius, its surface temperature will increase.
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.
A planet's orbital radius directly affects its orbital period through Kepler's third law of planetary motion. The farther a planet is from the star it orbits, the longer its orbital period will be, assuming all other factors remain constant. This relationship is expressed mathematically as T^2 ∝ r^3, where T is the orbital period and r is the orbital 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.
Doubling the mass of a satellite would result in no change in its orbital velocity. This is because the orbital velocity of a satellite only depends on the mass of the planet it is orbiting and the radius of its orbit, but not on the satellite's own mass.
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.
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.