Answer:
the <em>ratio F1/F2 = 1/2</em>
the <em>ratio a1/a2 = 1</em>
Explanation:
The force that both satellites experience is:
F1 = G M_e m1 / r² and
F2 = G M_e m2 / r²
where
- m1 is the mass of satellite 1
- m2 is the mass of satellite 2
- r is the orbital radius
- M_e is the mass of Earth
Therefore,
F1/F2 = [G M_e m1 / r²] / [G M_e m2 / r²]
F1/F2 = [G M_e m1 / r²] × [r² / G M_e m2]
F1/F2 = m1/m2
F1/F2 = 1000/2000
<em>F1/F2 = 1/2</em>
The other force that the two satellites experience is the centripetal force. Therefore,
F1c = m1 v² / r and
F2c = m2 v² / r
where
- m1 is the mass of satellite 1
- m2 is the mass of satellite 2
- v is the orbital velocity
- r is the orbital velocity
Thus,
a1 = v² / r ⇒ v² = r a1 and
a2 = v² / r ⇒ v² = r a2
Therefore,
F1c = m1 a1 r / r = m1 a1
F2c = m2 a2 r / r = m2 a2
In order for the satellites to stay in orbit, the gravitational force must equal the centripetal force. Thus,
F1 = F1c
G M_e m1 / r² = m1 a1
a1 = G M_e / r²
also
a2 = G M_e / r²
Thus,
a1/a2 = [G M_e / r²] / [G M_e / r²]
<em>a1/a2 = 1</em>
Answer:
The larger star will burn out faster. as nuclear fuel is burned out due to higher percentage being consumed.
Explanation:
Gravity on the other hand is only affected a little during this process and mass use. Solar mass loss is caused by mass that is turned into energy (E=mc²), but also by the material that is lost to the solar wind.
B. Reversing the current direction will cause the force deflecting the
wire to be perpendicular to the magnetic field but in the opposite
direction.
If you dropped a ball from any height, and measured its distance from the ground at any regular interval while it's falling, the graph of that distance versus time would be a graph that curves downward.
-- The ball is falling down. As time goes on, it gets closer and closer to the ground. Its remaining distance from the ground keeps decreasing, so the line on the graph slopes down.
-- The speed of the ball keeps increasing (it accelerates) because of the gravitational force on it. As time goes on, it covers more of the remaining distance during each interval than it did in the previous interval. The downward slope of the graph keeps increasing.
