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expeople1 [14]
3 years ago
14

If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa

ce of the earth
Physics
1 answer:
shepuryov [24]3 years ago
3 0

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

g=G\frac{m}{r^{2}}

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

g_{E}=G\frac{m}{r_{E}^{2}}

g_{P}=G\frac{m}{r_{P}^{2}}

The problem tells us the radius of the planet is twice that of the radius on earth, so:

r_{P}=2r_{E}

If we substituted that into the gravity of the planet equation we would end up with the following formula:

g_{P}=G\frac{m}{(2r_{E})^{2}}

Which yields:

g_{P}=G\frac{m}{4r_{E}^{2}}

So we can now compare the two gravities:

\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}

When simplifying the ratio we end up with:

\frac{g_{P}}{g_{E}}=\frac{1}{4}

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

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The gravitational force experienced by Earth due to the Moon is ________ the gravitational force experienced by the Moon due to
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Read 2 more answers
At locations A and B, the electric potential has the values VA = 1.83 V and VB = 5.17 V, respectively. A proton released from re
densk [106]

Answer:

a. It starts at point B.

vp = 2.53*10⁴ m/s

a. it starts at point A.

ve= 1.08*10⁶ m/s

Explanation:

a)  As the proton is a positive charge, when released from rest, it will be accelerated due to the potential difference, from the higher potential to the lower one, so it is at the point B when released.

Once released, as the total energy must be conserved, the increase in kinetic energy must be equal (in magnitude) to the change in the electric potential energy, as follows:

ΔK + ΔUe = 0 ⇒ ΔK = -ΔUe =- (e*ΔV)

⇒ -( e* (VA-VB) ) = \frac{1}{2}*mp*v^{2}

where e= elementary charge= 1.6*10⁻¹⁹ C,  VA = 1.83 V, VB= 5.17V, and mp= mass of proton = 1.67*10⁻²⁷ kg.

Replacing by these values, and solving for v, we have:

v = \sqrt{\frac{2*1.6e-19C*3.34 V}{1.67e-27kg} } = 2.53e4 m/s

⇒ vp = 2.53*10⁴ m/s

b) If, instead of a proton, the charge realeased from rest, had been an electron, a few things would change:

First, as the electrons carry negative charges, they move from the lower potentials to the higher ones, which means that it would have started at point A.

Second, as its charge is (-e) the change in electric potential energy had been negative also:

ΔUe = -e*ΔV = -e* (VB-VA)

In order to find the speed of the electron when it is just passing point B, we can apply the conservation of energy principle as for the proton, as follows:

-( (-e)* (VB-VA) ) = \frac{1}{2}*me*v^{2}

where e= elementary charge= 1.6*10⁻¹⁹ C,  VA = 1.83 V, VB= 5.17V, and me= mass of electron = 9.1*10⁻³¹ kg.

Replacing by these values, and solving for v, we have:

v = \sqrt{\frac{2*1.6e-19C*3.34 V}{9.1e-31kg} } = 1.08e6 m/s

⇒ ve = 1.08*10⁶ m/s

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