help When a falling meteoroid is at a distance above the Earth\'s surface of 3.4

Question

help

When a falling meteoroid is at a distance above the Earth's surface of 3.45 times the Earth's radius, what is its acceleration due to the Earth's gravitation? The motion of the meteoroid, or its mass, makes no difference. The gravitational field, which is the acceleration, will be smaller than the surface value we are used to because of the greater distance. The Earth exerts its force on the meteoroid according to mg = GmMe/r2 so we see that acceleration of gravity, g = GMe/r2 , follows an inverse-square law. At the surface, a distance of one Earth-radius (Ke) from the center, it is 9.80 m/s2. An altitude of 3.45 Re above the surface represents a distance of 4.45 Re from the center of the Earth, so the acceleration of gravity will be (4.45)2 = 19.80 times smaller.

Explanation / Answer

g = GMe/(4.45 Re)^2 = Gm/(4.45^2) (Re^2) = 9.8/19.8=0.495 m/s^2

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.