f=7,34 x10-7 ra_ Gm loi67408 x10-"0 yso.9.8 9.9.8 0. As you go above the Earth's surface, the acceleration due to gravity will decrease. Find the height, in meters, above the Earth's surface where this value will be 1/450 g. Where g =9.8 is the acceleration due to gravity on the Earth's surface. still can earth (Enter Ans: 1.29 X10^8 up to three significant figures using exponential notation, e g 4.29e+08 ) (F) converts to mg f=GMm %3D .2 2m on both sides so cross %3D r2 Mg: mg= GMMA we are looking put %3D .2 89.8 for r? crossoff move ra to other ト2 r2xg-GM clo this to %3D other 450 side. and get r but ř=Rth hais final answer side
Gravitational force
In nature, every object is attracted by every other object. This phenomenon is called gravity. The force associated with gravity is called gravitational force. The gravitational force is the weakest force that exists in nature. The gravitational force is always attractive.
Acceleration Due to Gravity
In fundamental physics, gravity or gravitational force is the universal attractive force acting between all the matters that exist or exhibit. It is the weakest known force. Therefore no internal changes in an object occurs due to this force. On the other hand, it has control over the trajectories of bodies in the solar system and in the universe due to its vast scope and universal action. The free fall of objects on Earth and the motions of celestial bodies, according to Newton, are both determined by the same force. It was Newton who put forward that the moon is held by a strong attractive force exerted by the Earth which makes it revolve in a straight line. He was sure that this force is similar to the downward force which Earth exerts on all the objects on it.
Attached is the question. He gave us the equation. I am just confused on where to find "M" earth mass?


Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images









