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Let's say we have two masses m_{0} for the earth and m_{1} for the moon. Then m_{0} does not equal m_{1}. Since the gravitational potential is Gm/r then at an identical radial distance from either source, the earth or the moon, we will obtain potentials that depend upon radial distance and mass. These will be different for each object. Showing that the curvature depends upon the amount of mass. If you argue against this then you show yourself to be ignorant of all scientific observations on the matter. In which case you should be listening to the wisdom of others.

Sun rays have to use more force to rotate the Planet. Obviously if the Gravity is strong, Gravity rays moving from the Planet interacts with Sun rays easily and even simple push makes the planet to rotate.

We are simply calculating Gravity using Newton’s “Universal law of Gravitation” which is not so correct.

but a planet having more mass loses Gravity quickly than a small planet.

Quote from: jeffreyH on 31/03/2018 14:25:14Let's say we have two masses m_{0} for the earth and m_{1} for the moon. Then m_{0} does not equal m_{1}. Since the gravitational potential is Gm/r then at an identical radial distance from either source, the earth or the moon, we will obtain potentials that depend upon radial distance and mass. These will be different for each object. Showing that the curvature depends upon the amount of mass. If you argue against this then you show yourself to be ignorant of all scientific observations on the matter. In which case you should be listening to the wisdom of others.It is my understanding that there is no gravity curvature , the object in orbit is travelling a linear path but forced to follow a curved path by the force of G to retain an equal radius from the sphere? The curve created relative to a constant velocity while gravity retains its linear hold. speed.jpg (20.13 kB . 740x464 - viewed 848 times)

Quote from: Thebox on 01/04/2018 20:31:54Quote from: jeffreyH on 31/03/2018 14:25:14Let's say we have two masses m_{0} for the earth and m_{1} for the moon. Then m_{0} does not equal m_{1}. Since the gravitational potential is Gm/r then at an identical radial distance from either source, the earth or the moon, we will obtain potentials that depend upon radial distance and mass. These will be different for each object. Showing that the curvature depends upon the amount of mass. If you argue against this then you show yourself to be ignorant of all scientific observations on the matter. In which case you should be listening to the wisdom of others.It is my understanding that there is no gravity curvature , the object in orbit is travelling a linear path but forced to follow a curved path by the force of G to retain an equal radius from the sphere? The curve created relative to a constant velocity while gravity retains its linear hold. speed.jpg (20.13 kB . 740x464 - viewed 848 times)So how would you describe the path it is following? What about the dilation of time at different radial distances from the source? Do you actually understand what curvature means in these circumstances?