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3 years ago

5

a man weighs 600 n while on the surface of earth. if he is transported to the planet mythos which has the same mass as earth but

a radius tjat os five times larger than earths , his weight would be

1 answer:

faltersainse [42]3 years ago

3 0

Answer:

24N

Explanation:

Calculation for what his weight would be

Let his WEIGHT on the surface of earth be 600 and Let the RADIUS be 25 which is (5^²) because we were told that the radius is FIVE TIMES larger than earths

Now let calculate his weight using this formula

Weight=Weight on Earth/Radius

Let plug in the formula

Weight=600N/5^²

Weight=600N/25

Weight=24N

Therefore his weight would be 24N

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Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1

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Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

$\sum F=ma$

We know that the force between two bodies due to gravity is given by the following equation:

$F_{g} = G\frac{m_{1}m_{2}}{r^{2}}$

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

$F_{g} =G\frac{Mm}{r^{2}}$

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

$F_{c}=ma_{c}$

where the centripetal acceleration is given by:

$a_{c}=\omega ^{2}r$

where

$\omega = \frac{2\pi}{T}$

Where T is the period, and $\omega$ is the angular speed of the planet, so:

$a_{c} = ( \frac{2\pi}{T})^{2}r$

or:

$a_{c}=\frac{4\pi^{2}r}{T^{2}}$

so:

$F_{c}=m(\frac{4\pi^{2}r}{T^{2}})$

so now we can do the sum of forces:

$\sum F=ma$

$F_{g}=ma_{c}$

$G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})$

in this case we can get rid of the mass of the planet, so we get:

$G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})$

we can now solve this for $T^{2}$ so we get:

$T^{2} = \frac{4\pi ^{2}r^{3}}{GM}$

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

$T_{1}^{2}>T_{2}^{2}$

So let's see what's going on there, we'll call:

$M_{1}$ = mass of Star 1

$M_{2}$ = mass of Star 2

So:

$\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}$

we can get rid of all the constants so we end up with:

$\frac{1}{M_{1}}>\frac{1}{M_{2}}$

and let's flip the inequality, so we get:

$M_{2}>M_{1}$

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

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3 years ago

Certain neutron stars (extremely dense stars) are believed to be rotating at about 500 rev/s. If such a star has a radius of 17

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Answer:

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So, F = gravitational force = GMm/R² where M = mass of neutron star, m = mass of object and R = radius of neutron star = 17 km

The centripetal force F' = mRω² where R = radius of neutron star and ω = angular speed of neutron star

So, since F = F'

GMm/R² = mRω²

GM = R³ω²

M = R³ω²/G

Since ω = 500 rev/s = 500 × 2π rad/s = 1000π rad/s = 3141.6 rad/s = 3.142 × 10³ rad/s and r = 17 km = 17 × 10³ m and G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²

Substituting the values of the variables into M, we have

M = R³ω²/G

M = (17 × 10³ m)³(3.142 × 10³ rad/s)²/6.67 × 10⁻¹¹ Nm²/kg²

M = 4913 × 10⁹ m³ × 9.872 × 10⁶ rad²/s²/6.67 × 10⁻¹¹ Nm²/kg²

M = 48,501.942 × 10¹⁵ m³rad²/s² ÷ 6.67 × 10⁻¹¹ Nm²/kg²

M = 7217.66 × 10²⁶ kg

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Answer:

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