Unit 6: Energy and Momentum of Rotating Systems
Weightage : 10-15%
Most-appropriate topic codes (AP Physics \(1\)):
(a)
The only force acting on the spacecraft is the gravitational force exerted by Earth. The arrow should start on the spacecraft and point toward the center of Earth.
\(\boxed{\text{Draw one force arrow toward Earth’s center, labeled }F_g\text{ or }F_{\text{Earth on spacecraft}}.}\)
(b)(i)
The gravitational force provides the centripetal force needed for circular motion.
\(F_g=F_c\)
\(\dfrac{GM_Em}{R^2}=\dfrac{mv^2}{R}\)
The spacecraft mass \(m\) cancels:
\(\dfrac{GM_E}{R^2}=\dfrac{v^2}{R}\)
\(v^2=\dfrac{GM_E}{R}\)
For circular motion, the orbital speed is
\(v=\dfrac{2\pi R}{T}\)
Substitute this into \(v^2=\dfrac{GM_E}{R}\):
\(\left(\dfrac{2\pi R}{T}\right)^2=\dfrac{GM_E}{R}\)
\(\dfrac{4\pi^2R^2}{T^2}=\dfrac{GM_E}{R}\)
\(T^2=\dfrac{4\pi^2R^3}{GM_E}\)
\(\boxed{T=\sqrt{\dfrac{4\pi^2R^3}{GM_E}}}\)
(b)(ii)
\(\boxed{\text{Equal to}}\)
The expression for orbital period is \(T=\sqrt{\dfrac{4\pi^2R^3}{GM_E}}\). It depends on the orbital radius \(R\), the mass of Earth \(M_E\), and the gravitational constant \(G\), but it does not depend on the spacecraft’s mass.
Therefore, a spacecraft of mass \(2m\) at the same orbital radius \(R\) has the same orbital period as the spacecraft of mass \(m\).
(c)
\(\boxed{\text{Less than}}\)
From the derivation in part (b)(i),
\(v^2=\dfrac{GM_E}{R}\)
so
\(v=\sqrt{\dfrac{GM_E}{R}}\)
This shows that orbital speed decreases as orbital radius increases. In the new orbit, the radius is greater than \(R\), so the spacecraft’s speed is less than its original speed.
\(\boxed{R\text{ increases } \Rightarrow v\text{ decreases}}\)
