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Korolek [52]
2 years ago
11

Which of the following numbers is correctly represented by 4 x 10-7? 0.0000004 40,000,000 0.000004

Physics
1 answer:
creativ13 [48]2 years ago
5 0
0.0000004. since it is 10^-7 (note the negative), you would move thedecimal point 7 places to the left.
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Calculate the force of gravity on the 0.60-kg mass if it were 1.3×107 m above Earth's surface (that is, if it were three Earth r
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The force of gravity between two objects is given by:
F=G \frac{m_1 m_2}{r^2}
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation

In this problem, the mass of the object is m_1=0.60 kg, while the Earth's mass is m_2=5.97 \cdot 10^{24} kg. Their separation is r=1.3 \cdot 10^7 m, therefore the gravitational force exerted on the object is
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lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A
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2.6×10^6\:\text{m}

Explanation:

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g = G\dfrac{M}{R^2}

and solving for R, we find that

R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration F_c experienced by the satellite is equal to the gravitational force F_G or

F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

v = \dfrac{2\pi r}{T}

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}

Solving for <em>M</em>, we get

M = \dfrac{4\pi^2 r^3}{GT^2}

Putting this expression back into Eqn(1), we get

R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}

\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}

\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}

\:\:\:\:= 2.6×10^6\:\text{m}

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