Finding gravitational acceleration near the surface of a planet with a satellite

AI Thread Summary
To find gravitational acceleration near a planet's surface using a satellite, the mass of the planet is calculated with the formula 4π²r³/t²G. This mass is then substituted into the universal gravitation equation Fg = GMm/R², which is set equal to mg. The distinction between using r and R in calculations hinges on whether the satellite is near the surface or at a higher altitude; r represents the satellite's distance from the planet's center, while R is the planet's radius. When the satellite is close to the surface, r can be approximated as R, simplifying the equation. Understanding these variables is crucial for accurate gravitational calculations.
katy123
Messages
1
Reaction score
0

Homework Statement


find acceleration due to gravity near the surface of a planet with a satellite by finding the mass of the planet and substituting it into the formula for the law of universal gravitation



Homework Equations


fc=mac
fg=GMm/R^2
fg=fc
fg=mg



The Attempt at a Solution


my question is a general one, I have solved for the mass of the planet finding that 4pi^2r^3/t^2G= Mass of planet
Then, you substitute it into Fg=GMm/R^2
set this equal to mg
mg=GMm/R^2

my teacher has done this on two separate occasions once that results in
acceleration due to gravity=4pi^2r/t^2
and another where it is = 4pi^2r^3/(R^2)(T^2)
r= distance from center to center
R=radius of the big planet
how do you know which radius to use in these equations, and why did it cancel out in one and not in the other?
 
Physics news on Phys.org
hi katy, try using the X2 and X2 buttons

It looks like in the first equation for the acceleration due to gravity your teacher approximated r ≈ R, which would be the case if you were on the surface of the Earth.

If you want to find the acceleration due to gravity on a satellite that is not near the surface of the earth, then r ≠ R.

r will just be the altitude of the satellite plus the radius of the Earth, which is R.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top