# Finding the minimum speed for putting an object in orbit

1. Mar 15, 2009

### pc2-brazil

yet another doubt about gravitation. we must affirm that it's also not a homework question, so, we won't need any number, only the formulas.

1. The problem statement, all variables and given/known data
1) suppose that an object is located in a distance r from the center of the Earth (not necessary to say that r is larger than the Earth's radius). we want to know the formula of the speed necessary (launching it horizontally) for that object to describe (A) a circular orbit of radius r.
we additionally want to know how to make this object describe (B) an elliptical, (C) a parabolic and a (D) hyperbolic orbit.
2) we would also like to know how to calculate the speed necessary to launch an object from the surface of the Earth, so it can enter in a circular orbit of radius r (bigger than the radius of the Earth).

2. Relevant equations
gravitational force:
Fg = GmM/r², where G is the gravitational constant, m is the mass of the object, M is the mass of the Earth and r is the distance between the center of the Earth and the center of the object (in this case, the distance r mentioned above).
centripetal force:
Fcp = mv²/r (where v is the speed of the object).
conservation of energy
$$\frac{mv^{2}_{0}}{2}-\frac{GmM}{R} = \frac{mv^{2}_{1}}{2}-\frac{GmM}{r}$$, where R is the radius of the Earth and r is the distance we mentioned above.

3. The attempt at a solution
1) (A) for the object to describe a circular orbit, its centripetal force has to balance with the gravitational force:
Fcp = Fg
mv²/r = GmM/r²
v²/r = GM/r²
v² = GM/r
$$v = \sqrt{\frac{GM}{r}}$$
for B, C and D, we have no clue, because we don't know the mathematical condition to obtain those orbits.
NOTE: does v have to be horizontal? why wouldn't an object thrown with vertical v = $$v = \sqrt{\frac{GM}{r}}$$ also respect that Fcp = Fg?
2) we don't have much idea where to start, but we think that this velocity will have a vertical component and a horizontal component which is $$v_x = \sqrt{\frac{GM}{r}}$$.
the vertical velocity would be v0y: $$\frac{mv^{2}_{0y}}{2}-\frac{GmM}{R} = \frac{mv^{2}_{1y}}{2}-\frac{GmM}{r}$$.
in this case, we think that v1y would have to be zero.