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Margarita [4]
3 years ago
5

Suppose we could shrink the Earth without changing its mass. At what fraction of its current radius would the free-fall accelera

tion at the surface be three times its present value?
Physics
2 answers:
spin [16.1K]3 years ago
7 0

Answer:

R' = \frac{1}{\sqrt{3}}R

Explanation:

The acceleration due to gravity on the surface of the Earth is given by:

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

where

G is the gravitational constant

M is the mass of the Earth

R is the radius of the Earth

Here we want to find the new Earth radius R' for which the gravitational acceleration at the surface, g', would be 3 times the current value of g:

g' = 3g

So we would have

\frac{GM}{R'^2}=3(\frac{GM}{R^2})

Solving the equation for R', we find

R'^2 = \frac{1}{3}R^2\\R' = \frac{1}{\sqrt{3}}R

musickatia [10]3 years ago
5 0

The earth can lose and gain weight.

<h2>Further Explanation </h2>

This may be a subject that is rarely discussed or even known by people. The interesting fact is that the Earth can actually lose mass and increase mass caused by several things. Basically every day is very possible for the Earth to collide with dust, comets, and meteors that can penetrate the atmosphere and land on Earth. This, of course, can indirectly increase its mass. On the other hand, mass loss occurs when there is a gas released from the atmosphere out so that the mass decreases when it occurs. Dr.a Chris Smith once tried to calculate this and it is estimated that each year 40,000 tons of space dust increases the mass of the Earth, but 95,000 tons of hydrogen and 1,600 tons of helium are released into space. On the other hand, it takes trillions of years for the hydrogen on this planet to run out but in contrast to helium which cannot be seen and is calculated in exact amounts. Some people have even warned that the Earth could run out of helium in the next three decades with the current conditions we are wasting on big projects.

If the size of the earth decreases, the force of gravity will decrease, the earth rotates faster, with lighter gravity, people will become lighter, faster, and have higher posture due to the lack of gravitational pressure.

Learn more

what if we shrink the earth  brainly.com/question/12644915

Details

Grade:  High School

Subject:  Physics

keywords: shrink, earth.

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aleksandrvk [35]

1) The forces are equal (Newton's third law of motion)

2) The force between the spheres will quadruple

3) The force of gravity exerted by the notebook on you is negligible

Explanation:

1)

In this part of the problem, we want to compare the gravitational force exerted by the larger mass sphere on the smaller mass sphere to the force exerted by the smaller mass sphere to the larger mass sphere.

We can do this by using Newton's third law of motion, which states that:

<em>"When an object A exerts a force (called </em><em>action</em><em>) on an object B, then object B exerts an equal and opposite force (called </em><em>reaction</em><em>) on object A"</em>

In this problem, we can identify the larger mass sphere as object A and the smaller mass sphere as object B. This law tells us that the two forces are equal in magnitude and opposite in direction: therefore, the gravitational force exerted by the larger mass sphere on the smaller mass sphere is equal to the force exerted by the smaller mass sphere to the larger mass sphere.

2)

The magnitude of the gravitational force between the two spheres is given by

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m_1, m_2 are the masses of the two spheres

r is the separation between the two spheres

In this problem, we are asked to find what happens when the distance between the spheres is halved, therefore when the new distance is

r'=\frac{r}{2}

Substituting into the equation, we find

F'=G\frac{m_1 m_2}{r'^2}=G\frac{m_1 m_2}{(r/2)^2}=4(\frac{Gm_1 m_2}{r^2})=4F

So, the force between the two spheres will quadruple.

3)

We can give an estimate for the gravitational force exerted by your notebook on you.

As we said, the magnitude of the gravitational force is

F=G\frac{m_1 m_2}{r^2}

Where:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

Let's estimate the following:

m_1 = 60 kg is your mass

m_2 = 2 kg is the mass of the notebook

r=1 m, assuming the notebook is at 1 metre from you

Substituting,

F=(6.67\cdot 10^{-11})\frac{(60)(2)}{1^2}=8.0\cdot 10^{-9} N

We see that this force has an extremely small value: therefore, it is almost negligible in daily life, where other much stronger forces act on you.

Learn more about gravity:

brainly.com/question/1724648

brainly.com/question/12785992

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