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3 years ago

Why does the moons gravity affect the earth so much

Physics

2 answers:

Licemer1 [7]3 years ago

7 0

One main reason is, The Moon's gravity pulls on Earth's oceans and distorts them, causing tides

Ronch [10]3 years ago

6 0

The Moon's gravity pulls on Earth's oceans and distorts them, causing tides. The water on the side of Earth closest to the Moon experiences the biggest pull, and bulges outward. The water on the opposite side also bulges, and the two bulges follow the Moon's motion and Earth's rotation.

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What two elements are named after Dimitri Mendeleev?

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4 years ago

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3 years ago

Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o

Juliette [100K]

(a) 0.473

The potential energy of a satellite orbiting around Earth is given by

$U=-\frac{GMm}{R+h}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

R is the Earth's radius

h is the altitude of the satellite above the Earth's surface

So the potential energy of satellite A is

$U_A=-\frac{GMm}{R+h_A}$

while potential energy of satellite B is

$U_B=-\frac{GMm}{R+h_B}$

Therefore the ratio of the potential energy of satellite B to that of satellite A is

$\frac{U_B}{U_A}=\frac{R+h_A}{R+h_B}$

and using

hA = 5920 km

hB = 19600 km

R = 6370 km

we find

$\frac{U_B}{U_A}=\frac{6370+5920}{6370+19600}=0.473$

(b) 0.473

The kinetic energy of a satellite orbiting around Earth instead is given by

$K=\frac{GMm}{2(R+h)}$

So the kinetic energy of satellite A is

$K_A=\frac{GMm}{2(R+h_A)}$

while kinetic energy of satellite B is

$K_B=\frac{GMm}{2(R+h_B)}$

Therefore the ratio of the kinetic energy of satellite B to that of satellite A is

$\frac{K_B}{K_A}=\frac{R+h_A}{R+h_B}$

which is identical to before, so it gives

$\frac{K_B}{K_A}=\frac{6370+5920}{6370+19600}=0.473$

(c) Satellite B

The total energy of a satellite in orbit is given by

$E=U+K = -\frac{GMm}{R+h}+\frac{GMm}{2(R+h)}=-\frac{GMm}{2(R+h)}$

We see that the total energy is:

1) negative (because the satellite is on a bound orbit)

2) inversely proportional to the distance of the satellite from the Earth's center, R+h

So the magnitude of the fraction in the equation is larger for the satellite which is closer to the Earth's surface (satellite A), but since the energy is negative, this means that the total energy of this satellite is smaller than that of satellite B. So, satellite B has a greater total energy.

(d) $1.03\cdot 10^7 J$