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.
a cubic foot of argon at 20 degrees celsius is isentropically compressed from 1 atm to 425 KPa. What is the new temperature and density?
Calculate the variance of the calculated accelerations. The free fall height was 1753 mm. The measured release and catch times were:
222.22 800.00
61.11 641.67
0.00 588.89
11.11 588.89
8.33 588.89
11.11 588.89
5.56 586.11
2.78 583.33
Give in the answer window the calculated repeated experiment variance in m/s2.
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Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning