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Margarita [4]
3 years ago
5

The mass of a planet is 3.7 x 1024 kg. If the planet has a radius of 9.2 x 106 m what is the acceleration of gravity for a perso

n on the surface of the planet? Find the escape velocity from the surface of this planet
Physics
1 answer:
nalin [4]3 years ago
3 0

Explanation:

It s given that,

Mass of a planet, M=3.7\times 10^{24}\ kg

Radius of a planet, R=9.2\times 10^{6}\ m

(1) We need to find the acceleration due to gravity for a person on the surface of the planet. Its formula is given by :

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

g=\dfrac{6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{(9.2\times 10^{6}\ m)^2}

g=2.91\ m/s^2

(2) The escape velocity is given by :

v=\sqrt{\dfrac{2GM}{R}}

v=\sqrt{{\dfrac{2\times 6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{9.2\times 10^{6}\ m}}

v = 7324.61 m/s

Hence, this is the required solution.

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A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o
blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}} (1)

Where:

T - Period of rotation of the satellite, measured in seconds.

r - Distance of the satellite with respect to the center of the planet, measured in meters.

G - Gravitational constant, measured in newton-square meters per square kilogram.

M - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}

r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} } (2)

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg and T = 25800\,s, then the distance of the satellite is:

r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }

r \approx 18.897\times 10^{6}\,m

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

F = \frac{G\cdot m\cdot M}{r^{2}} (3)

Where:

m - Mass of the satellite, measured in kilograms.

F - Force exerted on the satellite by the Earth, measured in newtons.

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg, m = 6105\,kg and r \approx 18.897\times 10^{6}\,m, then the gravitational force is:

F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}

F = 6841.905\,N

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0
3 years ago
What type of organization is used in a paragraph that lists similarities between two objects?
alekssr [168]
<span>A. Comparison 

</span>What type of organization is used in a paragraph that lists similarities between two objects? Comparison


NOT:
<span>B. Contrast 
C. Chronological order 
D. Cause and effect</span><span>
</span>
3 0
3 years ago
Most comets have short periods and orbit close to the ecliptic plane. (T/F)
Lapatulllka [165]

Answer:

False

Explanation:

Most comets are located outside the solar system, in part of the cloud that originated from dust and gas that has remained virtually untouchable for billions of years. The orbit of these comets can reach the order of a light year. Thus, they are called long-period comets.

3 0
3 years ago
A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertica
-BARSIC- [3]

Answer:

33.2 m

Explanation:

For the first object:

y₀ = 81.5 m

v₀ = 0 m/s

a = -9.8 m/s²

t₀ = 0 s

y = y₀ + v₀ t + ½ at²

y = 81.5 − 4.9t²

For the second object:

y₀ = 0 m

v₀ = 40.0 m/s

a = -9.8 m/s²

t₀ = 2.20 s

y = y₀ + v₀ t + ½ at²

y = 40(t−2.2) − 4.9(t−2.2)²

When they meet:

81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²

81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)

81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716

81.5 = 61.56t − 111.716

193.216 = 61.56t

t = 3.139

The position at that time is:

y = 81.5 − 4.9(3.139)²

y = 33.2

7 0
3 years ago
Susan gently pushes the tip of her finger against the eraser on her pencil and the pencil does not move. Which of the
pochemuha

Answer: D. ➡️⬅️

Explanation: I just knew the answer ;)

6 0
2 years ago
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