Question

As a body is projected to a high altitude above the earths surface, the variation of the acceleration of gravity with respect to altitude y must be taken into account. Neglecting air resistance, this acceleration is determined from the formula a=?g0[R2/(R+y)2], where g0 = 9.81 m/s2 is the constant gravitational acceleration at sea level, R = 6356 km is the radius of the earth, and the positive direction is measured upward.

With what velocity does the particle strike the earth if it is released from rest at an altitude y0 = 450 km?

Answer 1

The velocity of a particle projected to a high altitude can be calculated by considering the variation of the acceleration of gravity with respect to altitude. Neglecting air resistance, the acceleration of the particle is determined by the formula a=g0[R2/(R+y)2], where g0 = 9.81 m/s2 is the constant gravitational acceleration at sea level, R = 6356 km is the radius of the earth, and the positive direction is measured upward.

When released from rest at an altitude of y0 = 450 km, the particle will accelerate at a rate of 8.11 m/s2. Using the kinematic equations, we can calculate the velocity at which the particle will strike the earth's surface.

The equation for the velocity of the particle when it reaches the earth's surface is given by:

v = √2as

where v is the velocity, a is the acceleration, and s is the displacement.

Therefore, the velocity of the particle when it reaches the earth's surface is given by:

v = √2(8.11 m/s2)(6356 km - 450 km) = 20,680 m/s.

Hence, the velocity of the particle when it strikes the earth's surface after being released from rest at an altitude of 450 km is 20,680 m/s.