What Is the Acceleration at Point A Above a Planet?

[Karol]

Homework Statement

A mass m=100[kg] is released from stand still at point A which is in a distance of 1E7[m] from the face of a planet.
The planet's radius is also 1E7[m] and the free fall acceleration on it's face is 10[m/s².
What is the acceleration at point A

Homework Equations

The acceleration: $g=\frac{GM}{r^2}$

The Attempt at a Solution

First i find the mass:
$$10=\frac{6.7E-11\cdot M}{1E7^2}\Rightarrow M=1.49E18$$
$$g=\frac{6.7E-11\cdot 1.49E18}{2E7^2}=250E-9$$
Obviously this is wrong, the answer should be g=2.5

 

[TSny]

You have it set up correctly. Check your calculation of M. Did you treat (1E7)^2 correctly?

What about units, especially in your answer?

 

[adjacent]

$$g=\frac{6.7E-11\cdot 1.49E18}{2E7^2}=250E-9$$

You have got M correct but this part is wrong.

The radius is $(2 \times(10^7))^2$ not $2 \times (10^7)^2$

And yes, it's more accurate to treat $G$ as $6.67 \times 10^7$

 

[Karol]

I fixed it, i calculated wrong.
There is a harder continuation to this problem.
Approximate the time of fall by imposing upper and lower limits on the time.
I took the acceleration on the face and it gave a time of 1414[sec]. the upper limit is more complicated since the result is 2828[sec], and it isn't with the acceleration at A.
I guess i could approximate the acceleration to vary linearly with r: g=kr, but then i don't know to treat this kinematically

 

[TSny]

Homework Equations

The acceleration: $g=\frac{GM}{r^2}$

If you note that g varies as the inverse square of the distance, you should be able to get the answer for g at A without pencil and paper. What happens to g if r is doubled?

 

[TSny]

I took the acceleration on the face and it gave a time of 1414[sec]. the upper limit is more complicated since the result is 2828[sec], and it isn't with the acceleration at A.

Are you sure that you don't get 2828 s if you use the acceleration at A?

 

[dean barry]

r = planet radius = 1 E+7 metres
R = elevation = 1 E+7 metres
G = 6.7 E-11 (a constant)

Planet mass (M) from :
M = ( r ² * 10 ) / G
M = 1.493 E+25 kg

Only if you disregard the mass of body 2 (100 kg) as negligible (which it is, compared to the planet) can you simply calculate the field strength (g) from the basic equation :

g1 / g2 = ( ( r2 / r1 ) ² )
Key :
g1 = surface field strength ( ( m / s ) / s )
g2 = outer field strength ( ( m / s ) / s )
d1 = surface distance (m)
d2 = outer distance (m)

Transpose for g2 and calculate :
g2 = 2.5 ( ( m / s ) / s )

Your right with the time parameters, putting g as 2.5 and 10 gives the fall times of 2828 and 1414 seconds