Aspacecraft of mass 1.00×103kg orbits the sun (mass 1.99×1030kg) in a circular orbit of radius 1.50×1011m (equal to the average distance from the sun to the earth). you wish to move the spacecraft into a smaller circular orbit around the sun of radius 1.08×1011m (equal to the average distance from the sun to venus). in doing this, what will be the change in the spacecraft's kinetic energy? neglect the gravitational pulls of the planets on the spacecraft.

anchor = archer //

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft