Question

A 910-kg object is released from rest at an altitude of 1200 km above the north pole of the Earth. If we ignore atmospheric friction, with what speed does the object strike the surface of the earth? (G = 6.67 × 10-11 N · m2/kg2, MEarth = 5.97 × 1024 kg, the polar radius of the earth is 6357 km)

Answer 1

The speed of a 910-kg object striking the Earth after falling from 1200 km above the North Pole can be found using conservation of mechanical energy and the formula for gravitational potential energy.

Step-by-step explanation:

To determine the speed at which a 910-kg object strikes the surface of the Earth when released from rest at an altitude of 1200 km above the North Pole, we can use the conservation of mechanical energy. The total mechanical energy (kinetic plus potential) at the initial and final points must be equal since no external work is done on the system (we ignore atmospheric friction).

Initially, the object has potential energy due to its altitude above Earth and no kinetic energy since it's at rest. When it strikes the Earth, it has kinetic energy and minimal potential energy. Setting the initial and final total energies equal gives:

Ui + Ki = Uf + Kf,

where U is potential energy and K is kinetic energy. The potential energy can be calculated using the formula U = -GMEarthm/r, where m is the object's mass, r is the distance from the object to the center of Earth, and G is the gravitational constant. The kinetic energy is given by K = (1/2)mv2, where v is the velocity of the object.

The initial distance ri to the center of Earth is 6357 km + 1200 km (release altitude), and when the object strikes Earth, rf = 6357 km (polar radius of Earth). We know Ki = 0 and Uf is small enough to be negligible at the Earth's surface.

By plugging in the values, solving the equation for v, and taking the square root, we find the speed of the object when it strikes the ground.

Answer 2

In order to develop this problem it is necessary to use the concepts related to the conservation of both potential cinematic as gravitational energy,

`KE = \fract{1}{2}mv^2`

`PE = GMm((1)/(r_1)-((1)/(r_2)))`

Where,

M = Mass of Earth

m = Mass of Object

v = Velocity

r = Radius

G = Gravitational universal constant

Our values are given as,

`m = 910 Kg`

`r_1 = 1200 + 6371 km = 7571km`

`r_2 = 6371 km,`

Replacing we have,

`(1)/(2) mv^2 =  -GMm((1)/(r_1)-(1)/(r_2))`

`v^2 =  -2GM((1)/(r_1)-(1)/(r_2))`

`v^2 = -2*(6.673 *10^-11)(5.98 *10^24) ((1)/((7.571 *10^6)) -(1)/((6.371 *10^6)))`

`v = 4456 m/s`

Therefore the speed of the object when striking the surface of earth is 4456 m/s