Question

At an altitude of 1.3x10^7 m above the surface of the earth an incoming meteor mass of 1x10^6 kg has a speed of 6.5x10^3 m/s. What would be the speed just before impact with the surface of earth?Ignore air resistance.

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Answer 1

Approximately
`1.1 * 10^(4)\; {\rm m\cdot s^(-1)}` if air friction is negligible.

Step-by-step explanation:

Let
`G` denote the gravitational cosntant. Let
`M` denote the mass of the earth. Lookup the value of both values:
`G \approx 6.67 * 10^(-11)\; {\rm N\cdot m^(2)\cdot kg^(-2)}` while
`M \approx 5.697 * 10^(24)\; {\rm kg}`.

Let
`m` denote the mass of the meteor.

Let
`v_(0)` denote the initial velocity of the meteor. Let
`r_(0)` denote the initial distance between the meteor and the center of the earth.

Let
`r_(1)` denote the distance between the meteor and the center of the earth just before the meteor lands.

Let
`v_(1)` denote the velocity of the meteor just before landing.

The radius of planet earth is approximately
`6.371 * 10^(6)\; {\rm m}`. Therefore:

• At an altitude of
  `1.3 * 10^(7)\; {\rm m}` about the surface of the earth, the meteor would be approximately
  `r_(0) \approx 6.371 * 10^(6)\; {\rm m} + 1.3 * 10^(7)\; {\rm m} \approx 1.9 * 10^(7)\; {\rm m}` away from the surface of planet earth.
• The meteor would be only
  `r_(1) \approx 6.371 * 10^(6)\; {\rm m}` away from the center of planet earth just before landing.

Note the significant difference between the two distances. Thus, the gravitational field strength (and hence acceleration of the meteor) would likely have changed significant during the descent. Thus, SUVAT equations would not be appropriate.

During the descent, gravitational potential energy (
`\text{GPE}`) of the meteor was turned into the kinetic energy (
`\text{KE}`) of the meteor. Make use of conservation of energy to find the velocity of the meteor just before landing.

Initial
`\text{KE}` of the meteor:

`\displaystyle (\text{Initial KE}) = (1)/(2)\, m\, {v_(0)}^(2)`.

Initial
`\text{GPE}` of the meteor:

`\displaystyle (\text{Initial GPE}) &= -(G\, M\, m)/(r_(0))`.

(Note the negative sign in front of the fraction.)

Just before landing, the
`\text{KE}` and the
`\text{GPE}` of this meteor would be:

`\displaystyle (\text{Final KE}) = (1)/(2)\, m\, {v_(1)}^(2)`.

`\displaystyle (\text{Final GPE}) &= -(G\, M\, m)/(r_(1))`.
If the air friction on this meteor is negligible, then by the conservation of mechanical energy:

`\begin{aligned}& (\text{Initial KE}) + (\text{Initial GPE}) \\ =\; & (\text{Final KE}) + (\text{Final GPE})\end{aligned}`.

`\begin{aligned}& (1)/(2)\, m\, {v_(0)}^(2) - (G\, M\, m)/(r_(0)) \\ =\; & (1)/(2)\, m\, {v_(1)}^(2) - (G\, M\, m)/(r_(1))\end{aligned}`.

Rearrange and solve for
`v_(1)`, the velocity of the meteor just before landing:

`\begin{aligned}{v_(1)} &= \sqrt{\frac{\displaystyle (1)/(2)\, m\, {v_(0)}^(2) - (G\, M\, m)/(r_(0)) + (G\, M\, m)/(r_(1))}{(1/2)\, m}} \\ &= \sqrt{{v_(0)}^(2) - (G\, M)/(r_(0)) + (G\, M)/(r_(1))} \\ &= \sqrt{{v_(0)}^(2) - G\, M\, \left((1)/(r_(1)) - (1)/(r_(0))\right)}\end{aligned}`.

Substitute in the values and evaluate:

`\begin{aligned}v_(1) &= \sqrt{{v_(0)}^(2) - G\, M\, \left((1)/(r_(1)) - (1)/(r_(0))\right)} \\ &\approx \sqrt{\begin{aligned}(& 6.5 * 10^(3)\; {\rm m \cdot s^(-1)}) \\ & - [6.67 * 10^(-11)\; {\rm N \cdot {m}^(2)\cdot {kg}^(2) * 5.697\; {\rm kg}}\\ &\quad\quad * (1 / (6.371 * 10^(6)\; {\rm m}) - 1 / (1.9371 * 10^(7)\; {\rm m}))]\end{aligned}} \\ &\approx 1.1 * 10^(4)\; {\rm m\cdot {s}^(-1)}\end{aligned}`.

(Note that assuming a constant acceleration of
`g = 9.81\; {\rm m\cdot s^(-2)}` would give
`v_(1) \approx 1.7* 10^(4)\; {\rm m\cdot s^(-1)}`, an inaccurate approximation.