Question

A distant large asteroid is detected that might pose a threat to Earth. If it were to continue moving in a straight line at constant speed, it would pass 24000 km from the center of Earth. However, it will be attracted to Earth and might hit our planet. What is the minimum speed the asteroid should have so it will just graze the surface of the Earth?

Answer 1

Answer:

The minimum speed required is 5.7395km/s.

Explanation:

To escape earth, the kinetic energy of the asteroid must be greater or equal to its gravitational potential energy:

$K.E\geq P.E$

or

$\dfrac{1}{2}mv^2 \geq G\dfrac{Mm}{R}$

where $m$ is the mass of the asteroid, $R= 24,000,000\:m$ is its distance form earth's center, $M = 5.9*10^{24}kg$ is the mass of the earth, and $G = 6.7*10^{-11}m^3/kg\: s^2$ is the gravitational constant.

Solving for $v$ we get:

$v \geq \sqrt{\dfrac{2GM}{R} }$

putting in numerical values gives

$v \geq \sqrt{\dfrac{2(6.7*10^{-11})(5.9*10^{24})}{(24,000,000)} }$

$\boxed{v\geq 5739.5m/s}$

in kilometers this is

$v\geq5.7395m/s.$

Hence, the minimum speed required is 5.7395km/s.

Answer 2

The minimum speed of asteroid to graze from the surface of Earth is 5787.91 m/s.

Given data:

The distance of passing from the center of Earth is, $d=24000 \;\rm km =24 \times 10^{6} \;\rm m$.

The given problem is based on the concept of escape speed, which is the minimum speed of asteroid for grazing the surface of Earth. So, to escape from the surface of Earth, the kinetic energy of the asteroid must be greater or equal to its gravitational potential energy.

So,

kinetic energy = gravitational potential energy

$\dfrac{1}{2}mv^{2}=\dfrac{G \times M \times m}{d}$

Here,

M is the mass of Earth.

v is the minimum speed of Asteroid to escape from the Earth's surface.

G is the universal gravitational constant .

Solving as,

$v^{2}=2 \times\dfrac{G \times M}{d}\\\\v^{2}=2 \times\dfrac{6.7 \times 10^{-11} \times 6 \times 10^{24}}{(24\times 10^{6})^{2}}\\\\\\v=\sqrt{2 \times\dfrac{6.7 \times 10^{-11} \times 6 \times 10^{24}}{(24\times 10^{6})}}\\\\\\v=5787.91 \;\rm m/s$

Thus, we can conclude that the minimum speed of asteroid to graze from the surface of Earth is 5787.91 m/s.

Learn more about the escape speed here:

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