Un the way to the moon, the apollo astro-
nauts reach a point where the moon's gravi-
tational pull is stronger than that of earth's.
find the distance of this point from the
center of the earth. the masses of the
earth and the moon are 5.98 x 1024 kg and
7.36 x 1022 kg, respectively, and the distance
from the earth to the moon is 3.84 x 108 m.
answer in units of m.
020 (part 2 of 2) 10.0 points
what would the acceleration of the astro-
naut be due to the earth's gravity at this
point if the moon was not there? the
value of the universal gravitational constant
is 6.672 x 10-" n. m²/kg? .
answer in units of m/s? .

Distance =  345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

When we match these equations the masses cancel out as the universal gravitational constant

To solve this equation we have to replace the first equation of related with the distances.

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

Second part

The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

the independent variable, or the one that is intentionally manipulated is the speed.

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s