Anyone knows Gravity, and Plantetary Motion from physics class??

Question

A satellite is put into orbit at an altitude of 3x10^7 m above the ground. How fast will the satellite be moving? How do you figure out this problem?

Answer 1

First, use Newton's Law of Gravity to find the acceleration of the satellite...

a = G Me / r²

where G is the Newton Gravity Constant, Me the mass of the Earth, and r the distace from the center of the Earth to the satellite ( in your problem, 3 x 10^7 meters plus the radius of the Earth in meters ).

This acceleration is centripetal and is related to the tangential speed as your choice of

a = v² / r
a = r ω²

depending on whether or not you want v ( velocity in meters / second) or ω ( velocity in radians / second ).

Answer 2

It is known that:

a = (G*m)/(r^2) where:

a is the acceleration due to gravity
G is the gravitational constant 6.67*10^-11 N*(m^2)/(kg^2)
m is the mass of the object the gravity of which is calculated
r is the distance from the object the gravity of wihch is calculated

It is also known that:

a = (V^2)/r where:

a is centripetal acceleration
V is tangential velocity
r is the radius of the circular trajectory

In this problem the satellite's centripetal acceleration is equivalent to the pull of earth's gravity on it. Thus the expression:

V^2 = r*a is equivalent to the expression:

V^2 = (G*m)/r

r is the sum of the earth's radius and the altitude above ground: (3*10^7+6*10^6) = 4*10^7 m

The mass of the earth is 5.98*10^24 kg

V^2 thus equals [6.67*10^-11 N*(m^2)/(kg^2)]*[5.98*10^24]/[4*10^7 m] = 1*10^7 (m^2)/s^2

And V = sqrt(1*10^7 (m^2)/s^2) = 3*10^3 m/s.

The answer comes out somewhat ugly but that is because there is only one significant figure for the altitude.

Answer 3

You have to figure out the point at which the object's centrifugal force matches the gravitational pull.

F_g = -GMm/r²

F_c = mω²r

Solve for F_g = -F_c

GM/r² = ω²r

ω = sqrt(GM/r³)

G is the universal gravitational constant.
M is the mass of the earth
r is the radius of the orbit (3e7 m)
ω is the rotational velocity of the object.

I'll leave it to the asker to actually plug in the values and do the arithmetic.

Remember, the question asks about the satellite's speed. This generally refers to linear speed, so the rotational speed will need to be convered to linear speed (v = ωr).

Answer 4

F=Gm1m2/R²
F = Centripetal force =mV²/R
m1 mass of earth = 6.0*10^24
G is gravitational constant .(6.67*10^-11)
m2 mass of sattelite
R is distance from center of earth

m2V²/R=Gm1m2/R²
V²/R=Gm1/R²
V=√ (Gm1/R)
V=√ (6.67*10^-11)(6*10^24)/(6.4*10^6 + 3*10^7)
V=2288.715774 m/s