An object is released from rest at an altitude h above the surface of the Earth. (a) Show that its speed at a distance r from the Earth’s center, where R E ≤ r ≤ R E + h , is v = 2 G M E ( 1 r − 1 R E + h ) (b) Assume the release altitude is 500 km. Perform the integral Δ t = ∫ i f d t = − ∫ i f d r v to find the time of fall as the object moves from the release point to the Earth’s surface. The negative sign appears because the object is moving opposite to the radial direction, so its speed is v = − dr / dt . Perform the integral numerically.
An object is released from rest at an altitude h above the surface of the Earth. (a) Show that its speed at a distance r from the Earth’s center, where R E ≤ r ≤ R E + h , is v = 2 G M E ( 1 r − 1 R E + h ) (b) Assume the release altitude is 500 km. Perform the integral Δ t = ∫ i f d t = − ∫ i f d r v to find the time of fall as the object moves from the release point to the Earth’s surface. The negative sign appears because the object is moving opposite to the radial direction, so its speed is v = − dr / dt . Perform the integral numerically.
Solution Summary: The author explains the Gravitational potential energy of an object at a distance r from the Earth's center.
An object is released from rest at an altitude h above the surface of the Earth. (a) Show that its speed at a distance r from the Earth’s center, where RE ≤ r ≤ RE + h, is
v
=
2
G
M
E
(
1
r
−
1
R
E
+
h
)
(b) Assume the release altitude is 500 km. Perform the integral
Δ
t
=
∫
i
f
d
t
=
−
∫
i
f
d
r
v
to find the time of fall as the object moves from the release point to the Earth’s surface. The negative sign appears because the object is moving opposite to the radial direction, so its speed is v = −dr/dt. Perform the integral numerically.
Consider the circuit shown in the figure. The battery has emf ε = 69 volts and negligible internal resistance. The inductance is L = 0.4 H and the resistances are R 1 = 12 Ω and R 2 = 9.0 Ω. Initially the switch S is open and no currents flow. Then the switch is closed. After leaving the switch closed for a very long time, it is opened again. Just after it is opened, what is the current in R 1?
A capacitor with a capacitance of C = 5.95×10−5 F is charged by connecting it to a 12.5 −V battery. The capacitor is then disconnected from the battery and connected across an inductor with an inductance of L = 1.55 H . At the time 2.35×10−2 s after the connection to the inductor is made, what is the current in the inductor? At that time, how much electrical energy is stored in the inductor?
Can someone help me with this question. Thanks.
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