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Minchanka [31]
2 years ago
11

An astronaut lands on a new, recently discovered planet in a different star system. The astronaut measures the acceleration due

to gravity on the planet to be 12m/s2, and the mass of the planet is measured to be 7. 5E23kg. What is the radius of the new planet?
Physics
1 answer:
Nezavi [6.7K]2 years ago
3 0

Answer:

The radius of the new planet is ~2.04 * 10⁶ m, or 2,041,752 m.

Explanation:

We can use Newton's Law of Universal Gravitation:

  • \displaystyle F_g=G\frac{Mm}{r^2}

Let's look at Newton's 2nd Law:

  • F=ma

We can set these equations equal to each other:

  • \displaystyle G\frac{Mm}{r^2} =ma

The mass of the second mass (astronaut) cancels out. We are left with:

  • \displaystyle G\frac{M}{r^2} =a

We are solving for the radius of the new planet, so we can rearrange the equation:

  • \displaystyle r=\sqrt{\frac{GM}{a} }

Substitute in our known values given in the problem (<u><em>G = 6.67 * 10⁻¹¹ </em></u><em> ; </em><u><em>M = 7.5 * 10²³</em></u><em> ; </em><u><em>a = 12</em></u>).

  • \displaystyle r =\sqrt{\frac{(6.67\times 10^{-11})(7.5 \times 10^{23}}{12} }
  • r=2.04 \times 10^6

The radius of the new planet is ~2.04 * 10⁶ m.

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Answer:

q=2.997\times 10^{-4}C

Sign-Negative

Explanation:

We are given that

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We have to find the magnitude and sign of  charge would have to be placed on a penny .

By newton's second law

\sum F_y=ma

\sum F_y=qE-mg

Substitute the values then we get

qE-mg=ma

Substitute the values then we get

q(100)-3\times 10^{-3}(9.8)=3\times 10^{-3}(0.19)

100q-29.4\times 10^{-3}=0.57\times 10^{-3}

100q=0.57\times 10^{-3}+29.4\times 10^{-3}=29.97\times 10^{-3}

q=\frac{29.97\times 10^{-3}}{100}

q=2.997\times 10^{-4}C

Sign of charge =Negative

Because electric force acting  in opposite direction of electric field therefore,charge on penny will be negative.

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