'acceleration due to gravity' comparing dissimiliar distances/acceleration for the same final speed?

Question

I'm 90,000 miles from a distant 'virtual' planet. Call it a point. Beginning from zero, I continue to accelerate. Passing through that point I record my speed. (the mass of the planet can be anything).

I do this again while increasing my distance from this point to 110,000 miles. I record my speed at exactly 100,000 miles. (10,000 miles short of the point).

Are they the same speeds? Can someone help me to set this problem up so I know what I'm doing in the future? I'm not a great algebra student. Yes, I tried astonomy/space.

eyeonthe... hi, there is, under some circumstance, an equality. can you explain this in simple laymans language?

Don S... does this take into account that the acceleration is less for the 'second' trip but longer trip?

Answer 1

No. The speeds are different.

A: -90,000 mi -> 0
B: -110,000 mi -> -10,000mi

Velocity of A at 0 will be greater than B at -10,000.
In fact, since mass is at a point, velocity of A at 0 → ∞
>>F = G*M*m/r^2 =a*m
>>a = F/m = H*M/r^2
>>as r→0, a→∞

At other distances, eg.
A: -100,000 mi -> -10,000mi
B: -110,000 mi -> -20,000mi
Velocity of A at -10,000 mi is still greater than
Velocity of B at -20,000 mi.
Because
acceleration A from -100,000mi to -10,000mi
is greater than
acceleration B from -110,000mi to -20,000mi

For setup where v_initial=0,
v_final = [2*G*M* (1/r_f - 1/r_i)]^0.5
G = gravitational constant
M = mass of big object
r_i = distance from center of big object, initial
r_f = distance from center of big object, final

Answer 2

Let's see if I have this straight. You're at r = 90,000 mi from a planet of mass M. Then you move on to R = 110,000 mi. At r + S = 100,000, you record your velocity v. S = ut + 1/2 at^2, which is the distance you traveled in time t while accelerating at a. u is the initial velocity you were at when at r.

v = u + at; where u is the initial velocity at r and v is the velocity you recorded at r + S. Thus, when a <> 0, v <> u and the answer is, no they are not the same velocities. If there were no average acceleration over the distance S, then v = u and they would be the same velocities.

Answer 3

Well your acceleration is a=GM/R^2. This comes from the gravitational force equation F=GmM/R^2. F=ma and your little m's cancel out. For your first run you are going to have a v1=GM/(90,000)^2*t1. Where t1 is the time it took you to get there. For your second run you are going to get v2=GM/(100,000)^2 *t2. t2 is going to be greater than t1 since your a is GM/R^2 and is more heavily dependent upon R. Therefore no your velocities will not be the same

Answer 4

no because you stated that you coninue to accerate. speed is velocity. v=x/t. acceleration is a=v/t.