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sveta [45]
2 years ago
6

What keeps all of the planets in the solar system from leaving their orbits around the sun?.

Physics
1 answer:
nadya68 [22]2 years ago
6 0
It’s the gravity pull of the sun that makes them spin around 360 degrees
It’s all so why they say In a billion years the sun will smash into the sun
-facts :)
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3 years ago
A m = 94.2 kg object is released from rest at a distance h = 1.15134 R above the Earth’s surface. The acceleration of gravity is
OleMash [197]

Answer:

v= 4055.08m/s

Explanation:

This is a problem that must be addressed through the laws of classical mechanics that concern Potential Gravitational Energy.

We know for definition that,

U = \frac{GMm}{r}

We must find the highest point and the lowest point to identify the change in energy, so

Point a)

The problem tells us that an object is dropped at a distance of h = 1.15134R over the earth.

That is to say that the energy of that object is equal to,

U_1=-\frac{(6.6738 * 10^{-11})(5.98 * 10^{24})(94.2)}{(1.15134)(6.38*10^6)}

U_2= - 5.1180*10^9J

Point B )

We now use the average radius distance from the earth.

U_2=-\frac{(6.6738 * 10^{-11})(5.98 * 10^{24})(94.2)}{(6.38*10^6)}

U_2= -5.8925*10^9J

Then,

\Delta U = U_2 - U_1 = -5.1180*10^9J - ( -5.8925*10^9J)

\Delta U = 774.5*10^6

By the law of conservation of energy we know that,

\Delta U = \frac{1}{2}mv^2

clearing v,

v= \sqrt{2 \Delta U/m}

v= \sqrt{2*774.5*10^6 /94.2}

v= 4055.08m/s

Therefore the speed of the object when it strikes the Earth’s surface is 4055.08m/s

8 0
3 years ago
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