Question

when a falling meteoroid is at a distance above the earth's surface of 3.30 times the earth's radius, what is its acceleration due to the earth's gravitation?

Answer 1

Approximately
`0.515\; {\rm m\cdot s^(-2)}` (assuming that all other forces are negligible).

Step-by-step explanation:

Look up the following quantities:

• Gravitational constant:
  `G \approx 6.6743 * 10^(-11)\; {\rm m^(3)\cdot kg^(-1)\cdot s^(-2)}`.
• Mass of the Earth:
  `M \approx 5.792 * 10^(24)\; {\rm kg}`.
• Radius of the Earth:
  `R \approx 6.371 * 10^(6)\; {\rm m}`.
  (Ensure that this value is in the standard unit of meters.)

At a distance of
`r` from the center of the Earth, the magnitude of the gravitational field strength of the Earth would be:

`\begin{aligned} g &= (G\, M)/(r^(2))\end{aligned}`.

In this question, the asteroid is at
`3.30\, R` above the surface of the Earth. The actual distance between the asteroid and the center of the Earth would be
`r = (3.30 + 1)\, R`.

Thus, at the current position of this asteroid, the magnitude of the gravitational field of the Earth would be:

`\begin{aligned} g &= (G\, M)/(r^(2)) \\ &= (G\, M)/(((3.30 + 1)\, R)^(2)) \\ &\approx \frac{(6.6743 * 10^(-11)\; {\rm m^(3)\cdot kg^(-1)\cdot s^(-2)})\, (5.792 * 10^(24)\; {\rm kg})}{((3.30 + 1)\, (6.371 * 10^(6)\; {\rm m}))^(2)} \\ &\approx 0.515\; {\rm m\cdot s^(-2)}\end{aligned}`.

Assume that the gravitational attraction from the Earth is the only force on this asteroid. Acceleration of the asteroid would be equal to the gravitational field strength: approximately
`0.515\; {\rm m\cdot s^(-2)}`.