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

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

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

$PE = GMm(\frac{1}{r_1}-(\frac{1}{r_2}))$

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,

$\frac{1}{2} mv^2 =  -GMm(\frac{1}{r_1}-\frac{1}{r_2})$

$v^2 =  -2GM(\frac{1}{r_1}-\frac{1}{r_2})$

$v^2 = -2*(6.673 *10^-11)(5.98 *10^24) (\frac{1}{(7.571 *10^6)} -\frac{1}{(6.371 *10^6)})$

$v = 4456 m/s$

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

Answer 2

Energy can neither be created nor be destroyed. The velocity of the object when it will fall from an altitude of 1200 km is 1.41044 × 10⁵ m/s.

What is the law of conservation of energy?

The law of conservation of energy states that energy can neither be created nor be destroyed, it can only transform from one form to another.

As we know that the total energy of the system is always conserved, therefore, the kinetic energy of the ball will be equal to the potential energy of the system.

The kinetic energy of the system, $KE = \dfrac{1}{2}mv^2$

The potential cinematic energy of the system, $PE= GMm(\dfrac{1}{r_1}-\dfrac{1}{r_2})$

As it is given that the mass of the earth is  5.97 × 1024 kg, while the value of the gravitational constant is G = 6.67 × 10-11 N · m2/kg2, and the radius of the earth is given as 6375 km. Further, the radius of the ball can be written as (6357+1200 = 7557 km).

Now, equating the energy equations we will get,

$KE = PE\\\\\dfrac{1}{2}mv^2 = GMm(\dfrac{1}{r_1}-\dfrac{1}{r_2})\\\\\dfrac{1}{2}mv^2 = -GMm(\dfrac{1}{r_2}-\dfrac{1}{r_1})\\\\\dfrac{1}{2}v^2 = -GM(\dfrac{1}{r_2}-\dfrac{1}{r_1})\\\\v^2 = 2 \times [-GM(\dfrac{1}{r_2}-\dfrac{1}{r_1})]\\v^2 = 2 \times [-6.67 \times 10^{-11}\times 5.97 \times 10^{24}(\dfrac{1}{7557}-\dfrac{1}{6357})]\\\\v^2 = 1.9893\times 10^{10}\\\\v = 1.41044 \times 10^5\rm\ m/s$

Hence, the velocity of the object when it will fall from the altitude of 1200 km is 1.41044 × 10⁵ m/s.

Learn more about the Law of conservation of Energy:

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