The Earth's equator is an imaginary planetary line that is about 40,075 km (24,901 mi) long in circumference. The equator divides the planet into the Northern Hemisphere and Southern Hemisphere and is located at 0 degrees latitude, the halfway line between the North Pole and South Pole.In spatial (3D) geometry, as applied in astronomy, the equator of a rotating spheroid (such as a planet) is the parallel (circle of latitude) at which latitude is defined to be 0°. It is the imaginary line on the spheroid, equidistant from its poles, dividing it into northern and southern hemispheres. In other words, it is the intersection of the spheroid with the plane perpendicular to its axis of rotation and midway between its geographical poles.
When the Sun is directly above the Earth's equator (on the equinoxes of approximately March 20 and September 23), sunlight shines perpendicular to the Earth's axis of rotation, and all latitudes have a 12-hour day and 12-hour night. On and near the equator sunlight comes from almost directly above every day year-round, and thus the equator has a rather stable daytime temperature the whole year.
I am trying to find what velocity and angle of launch is required for a projectile to be fired from the North pole and land somewhere on the equator. I was thinking 45 degrees with muzzle velocity 9401m/s but that sounds ridiculous. Also how much time would it be in the air for?
according to Newtons law since the Earth is flatten in the poles the distance to the center of the Earth is bigger in the equator and therefore its gravity is stronger there.
with this same argument you could say that if you got to the center of the Earth the distance tends to 0 and therefore...