Weight = mass * acceleration due to gravity.
In the SI system, g = 9.8 m/s^2.
Hence, your weight at the equator will be 9.8xm Newton. ( I dont know your mass. you see. so I'm writing 'm')
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it would weigh the same because the mortor comes from earth
no
This would be known as the Equator
If you assume thatequatriol = equatorial raduis = radius then the equatorial radius is the radius of the spheroid measured at its equator. It would be = length of the equator/(2*pi)
The earth's equator is approximately 40,075 kilometres. There is no exact value because, amongst other things, the equator is not static: it moves with shifts in the axis of the earth's rotation.However, using calculus, it is possible to show that the length of the string would need to be 2*pi inches = 6.3 inches greater than the length of the equator measured in inches. Given the variability in measuring the earth's equator, that difference will not be identifiable.The earth's equator is approximately 40,075 kilometres. There is no exact value because, amongst other things, the equator is not static: it moves with shifts in the axis of the earth's rotation.However, using calculus, it is possible to show that the length of the string would need to be 2*pi inches = 6.3 inches greater than the length of the equator measured in inches. Given the variability in measuring the earth's equator, that difference will not be identifiable.The earth's equator is approximately 40,075 kilometres. There is no exact value because, amongst other things, the equator is not static: it moves with shifts in the axis of the earth's rotation.However, using calculus, it is possible to show that the length of the string would need to be 2*pi inches = 6.3 inches greater than the length of the equator measured in inches. Given the variability in measuring the earth's equator, that difference will not be identifiable.The earth's equator is approximately 40,075 kilometres. There is no exact value because, amongst other things, the equator is not static: it moves with shifts in the axis of the earth's rotation.However, using calculus, it is possible to show that the length of the string would need to be 2*pi inches = 6.3 inches greater than the length of the equator measured in inches. Given the variability in measuring the earth's equator, that difference will not be identifiable.