Where does the potential energy of rocket goes when it escape from the influence of the earth

1 answer:

6 0

The potential energy of a rocket increases when its escapes from the influence of the earth. It increases when it escapes the earth's boundary.

<h3>What is escape speed?</h3>

Escape velocity or escape speed is the minimum speed required for a free, non-propelled object to escape from the gravitational pull of the main body and reach an infinite distance from it in celestial physics.

It is commonly expressed as an ideal speed, neglecting atmospheric friction.

If a spacecraft is launched from the moon at the escape speed of the earth then the value of escape speed will be the difference in both the velocities.

An object's height above the zero position directly relates to its gravitational potential energy.

The kinetic energy of motion is converted into gravitational potential power as your rocket ascends.

The rocket possesses gravitational potential energy as it climbs because it is becoming farther away from the Earth's surface, much as a positive and negative charge moving apart.

The potential energy of a rocket increases when its escapes from the influence of the earth. It increases when it escapes the earth's boundary.

To learn more about the escape speed refer to the link;

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A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,

viktelen [127]

Part A)
First of all, let's convert the radii of the inner and the outer sphere:
$r_A = 11.0 cm = 0.110 m$
$r_B = 16.5 cm=0.165 m$
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
$C=4 \pi \epsilon _0 \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=$
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Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
$Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C$

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
$E\cdot (4 \pi r^2) = \frac{Q}{\epsilon _0}$
from which
$E(r) = \frac{Q}{4 \pi \epsilon_0 r^2}$
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
$E(0.111m)=2680 N/C$

And so, the energy density at r=0.111 m is
$U= \frac{1}{2} \epsilon _0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3$

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: $Q=3.67 \cdot 10^{-9}C$ . We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
$E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C$

And therefore, the energy density at this distance from the center is
$U= \frac{1}{2}\epsilon_0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3$

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