space question (1 Viewer)

[Xu]

"What is the free-fall acceleration of an object at the altitude of the space
shuttle's orbit, about 400 km above the earth's surface?"

can anyone help me solve this question? its odd numberd in texbook and no answer in back but I think I do wrong regaurdless.

thank you so much!

 

[KeypadSDM]

F = m_objecta
F = G.m_E.m_object/[r_Earth + 400 * 1000]²

Simplifying yields:
a = G.m_E/[r_Earth + 400 * 1000]²

EDIT: I hate the physics board. Just change all the things surrounded by subs to subscripts, and sups to superscripts.

 

[潘偉勤]

You should use Newton's Law of Gravitation.

F = ma = GmM/r^2

a = GM/r^2

r = radius of Earth + altitude = 6800 * 10^3 m
M = mass of Earth = 6 * 10^24 kg
G = Newton's gravitational constant = 6.67 * 10^-11 m^2 N·kg^-2

Bung them into the equation, and you should get a = 8.65 m·s^-2.

 

[Xu]

thank you both for shaving me from headake! it was easier than I thought first.
can I ask one more question? it seems really simply like before but I seem to had troable with it like just then. thank you!

"A satellite travels in a circular orbit around the earth. Find its period if the satellite is just above the surface of the earth."

no other infomation..

 

[潘偉勤]

Shaving you, huh? No problem.

This next question isn't as straightforward as the first. Again, you use the same formula.

F = GmM/r² = mv²/r

v = (GM/r)^1/2

G = 6.67 * 10^-11 m² N·kg^-2
r = 6400 * 10³ m
M = 6 * 10²⁴ kg

v = 7907.67 m·s^-1

For the period T, you use the speed-distance-time relationship.

T = d/v
d = 2πr

T = 2πr/v

So T = 5085.24 s = 1.41 hours.