Can someone help me derive this equation?

Question

how do I derive these equations into formulas for velocity, period, and radius of orbit for an object?
F=GMm/d^2, F=mv^2/r (centripetal) where v=2(3.14)r/T

Answer 1

For Velocity:
F = GMm / d²
Fd² = GMm
Fd²/ GM = m
But from the 2nd equation, m = Fr/ v², subsitute in the value.
Fd²/ GM = Fr/ v²
Fd²/ GMFr = 1/ v²
GMFr/ Fd² = v²
√[GMFr/ Fd²] = v
v = √[GMFr/ Fd²]

For Period:
F = GMm / d²
Fd² = GMm
Fd²/ GM = m
But from the 2nd equation, m = Fr/ v², subsitute in the value.
Fd²/ GM = Fr/ v²
Fd²/ GMFr = 1/ v²
GMFr/ Fd² = v²
√[GMFr/ Fd²] = v
v = √[GMFr/ Fd²]
But v = 2π r / T
2π r / T = √[GMFr/ Fd²]
2π r = T √[GMFr/ Fd²]
2π r / √[GMFr/ Fd²]= T
T = 2π r / √[GMFr/ Fd²]

For Radius:
F = GMm / d²
Fd² = GMm
d² = GMm/ F
d = √[GMm/ F]
2r = √[GMm/ F]
r = √[GMm/ F] / 2



For Velocity:
F = mv²/ r
Fr = mv²
Fr /m = v²
v² = Fr /m
v = √(Fr /m)

For Period:
F = mv²/ r
Fr = mv²
Fr /m = v²
v² = Fr /m
v = √(Fr /m)
2π r / T = √(Fr /m)
2π r = T√(Fr /m)
2π r / [√(Fr /m)] = T
T = 2π r / [√(Fr /m)]

For Radius:
F = mv²/ r
Fr = mv²
r = mv²/ F


For Velocity:
v = 2π r / T

For Period:
v = 2π r / T
vT = 2π r
T = 2π r/ v

For Radius:
v = 2π r / T
vT = 2π r
vT/ 2π = r
r = vT / 2π

Answer 2

Set the two forces equal to each other:

a) F = GMm/r^2 = mv^2/r

Divide m from both sides, then, substitute the velocity expression into a)

b) GM/r^2 = (4 pi^2 r^2/T^2)/r

Solve for T^2:

c) T^2 = [4 pi^2/(GM)] r^3

Answer 3

lets do the velocity one first hehe.

mv^2/r = GMm/d^2
*(note that the r on the left hand side refer to the radius of orbit which is the same as d on the right hand side of the equation)

cancelling out m and r we get
v^2 = GM/r
to get v we simply apply square root to GM/r

ok now we have velocity

for the period, T, just sub v=2(3.14)r/T into v^2 = GM/r and make T the subject.

Hope it helps :)

Answer 4

That's calculus basically. A derivative shows the instantaneous change of a function (so how fast it's changing at a point). It's given by lim as h--> 0 of (f(x+h) - f(x))/h The basics of it are, Power rule: derivative of x^a is ax^(a-1) Coefficent rule: derivative of a f(x) is a f'(x) (f'(x) denotes derivative of f(x)) Chain rule: f(g(x)) derived is by definition f'(g(x))g'(x) So like say you have ( x^2 + 3)^3 would be 3(x^2 + 3)^2 (2x) Hope that helps.

Answer 5

unless I'm missing something or made a mistake,
v=fr/m where the right side is under the root
r=mv^2/F

Answer 6

Energy =force x distance that the force travels
E=(GM1*M2/r^2) x radius of gravitational mass
If we divide the gravity Energy by the masss which is rotating around the gravitational mass we obtain the following formula;
E/M=V^2 =GM1/r
V=(GM1/r)^0.5
R=v x t =GM1/V^2
Tr=radial gravitational time=R/V
from this you can derive Keppler's formula.

Answer 7

for velocity
mv^2/r=GMm/r^2
v^2=GM/r
v=(GM/r)^1/2

for time period
v=distance/time
T=d/v
t=2*pi*r/(GM/r)^1/2
t={4*pi^2*r^3/GM}^1/2

for radius
r=GM/v^2