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Anon25 [30]
4 years ago
9

Derive an expression for the energy needed to launch an object from the surface of Earth to a height h above the surface.Ignorin

g Earth's rotation, how much energy is needed to get the same object into orbit at height h?Express your answer in terms of some or all of the variables h, mass of the object m, mass of Earth mE, its radius RE, and gravitational constant G.
Physics
1 answer:
Sergio [31]4 years ago
8 0

Answer:

Explanation:

Potential energy on the surface of the earth

= - GMm/ R

Potential at height h

=  - GMm/ (R+h)

Potential difference

= GMm/ R -  GMm/ (R+h)

= GMm ( 1/R - 1/ R+h )

= GMmh / R (R +h)

This will be the energy needed  to launch an object from the surface of Earth to a height h above the surface.

Extra  energy is needed to get the same object into orbit at height h

= Kinetic energy of the orbiting object at height h

= 1/2 x potential energy at height h

= 1/2 x GMm / ( R + h)

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sammy [17]

Hi, what is it you need help with?

8 0
2 years ago
Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject materia
melisa1 [442]

Answer:

H_2 = 91.55 km

Explanation:

Gravity on the surface of planet is given as

g = \frac{GM}{R^2}

as we know that

M = 8.93 \times 10^{22} kg

R = 1821 km

now gravity on the planet is

g = \frac{(6.67 \times 10^{-11})(8.93 \times 10^{22})}{(1821 \times 10^3)^}

so we have

g = 1.8 m/s^2

now we know that

H_{max} = \frac{v^2}{2g}

so we will say

\frac{H_1}{H_2} = \frac{g_2}{g_1}

\frac{500}{H_2} = \frac{9.81}{1.8}

H_2 = 91.55 km

5 0
3 years ago
At time t=0 a 2150 kg rocket in outer space fires an engine that exerts an increasing force on it in the +x−direction. This forc
amm1812

Explanation:

Given that,

Mass of the rocket, m = 2150 kg

At time t=0 a rocket in outer space fires an engine that exerts an increasing force on it in the +x−direction. The force is given by equation :

F=At^2

Here F = 888.93 N when t = 1.25 s

(c) We can find the value of A first as :

F=At^2\\\\A=\dfrac{F}{t^2}\\\\A=\dfrac{888.93}{(1.25)^2}\\\\A=568.91\ N/s^2

The value of A is 568.91\ N/s^2.

(a) Let J is the impulse does the engine exert on the rocket during the 4.0 s interval starting 2.00 s after the engine is fired. It is given in terms of force as :

J=\int\limits {F{\cdot} dt}

Limits will be from 2 s to 2+ 4 = 6 s

It implies :

J=\int\limits^6_2 {At^2{\cdot} dt}\\\\J=A\int\limits^6_2 {t^2{\cdot} dt}\\\\J=A\dfrac{t^3}{3}|_2^6\\\\J=568.91\times \dfrac{1}{3}\times (6^3-2^3)\\\\J=39444.42\ Ns

(b) Impulse is also equal to the change in momentum as :

J=m\Delta v\\\\\Delta v=\dfrac{J}{m}\\\\\Delta v=\dfrac{39444.42}{2150}\\\\\Delta v=18.34\ m/s

Hence, this is the required solution.

5 0
3 years ago
Which term best describes the process that occurs when light passes through a lens? A. reflection B. refraction C. resonance D.
Elza [17]
A. Refraction, meaning the same thing...it just passes through. Reflection is when light, heat, or sound bounces off a surface, kind of like an echo. Resonance is sound reverberating, or vibrating. And polarization is the action of not allowing the vibrations of a transverse wave. Hoped this helps you :) 

Short answer: Refraction
3 0
3 years ago
Read 2 more answers
Two similar pyramids have base areas of 12.2 cm2 and 16 cm2. the surface area of the larger pyramid is 56 cm2. what is the surfa
Jlenok [28]

The surface area of the smaller pyramid is determined as 42.7 cm².

<h3>Area of pyramids</h3>

The base area of the two similar pyramid is given as follows;

Base area of large pyramid = 16 cm²

Base area of smaller pyramid = 12.2 cm²

<h3>Surface area of the similar pyramids</h3>

The surface area of the smaller pyramid is calculated as follows;

s₁/b₁ = s₂/b₂

s₁/12.2 = 56/16

s₁ = 42.7 cm²

Thus, the surface area of the smaller pyramid is determined as 42.7 cm².

Learn more about base area of pyramids here: brainly.com/question/16060915

#SPJ4

6 0
3 years ago
Read 2 more answers
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