PLEASE HELP PHYSICS gravitation questions!!!?

Question

1. At what horizontal velocity would a satellite have to be launched from the top of Mt. Everest to be placed in a circular orbit around the Earth?

=.............m/s

2. During an Apollo lunar landing mission, the command module continued to orbit the Moon at an altitude of about 148 km. How long did it take to go around the Moon once?
=.........s

3. What is the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to the Earth is 1/38 of its value at the Earth's surface?
=.............m

Answer 1

1. Gee what a low flying satellite

Force up = Force down
m[ (Vh)^2]/R=mg
or
Vh=sqrt(Rg)
R is the distance from the center of the Earth to the launching pad on that Mt. Everest of yours. Are you sure you want to that? That is only 29,000 feet... might interfere with commercial aviation.

PS if you want to do a super job then replace F=mg with F=G M1m2/R^2 [G- universal gravitationa l constant]
m2[ (Vh)^2]/R=G M1m2/R^2
Vh=sqrt(G M1/R)

2. During an Apollo lunar landing mission... Oh yea. It seems just like yesterday.

Same math
V=sqrt(G M1/R)
R= 148,000 + R(moon)
M1 - mass of the moon (that is more than a few kg)

Finally considering a circular orbit

t=2 pi R/V


3. Is that a trick question? Hmmm...

F1=GM1m/R1^2 (that is on the surface)
F2=GM1m/R2^2 (that is above surface)

F2/F1=1/38=[GM1m/R2^2]/[GM1m/R1^2]=
1/38=[R1/R2]^2

R2= sqrt(38)R1 and finally

D=R1-R2

Answer 2

Your formula is somewhat incorrect. you may desire to divide by d², relatively than multiply. So, we ought to continually have... F = (Gm1m2)/d² the place G has the fee 6.673 x 10^-11 the awesome result will as a result be... F = (G x 8.a million x 6.5)/0.5² = 0.00000001405