Can you help me with this??

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

If you could go a year into the future, or 3 seconds into the past, where would you go? What date? I would go to June 6th, 2018

vova2212 [387]

Idk I would go back to pre-k

8 0

3 years ago

A leaf falls from a tree. Because the leaf is so light, it will be affected by air

postnew [5]

Answer:

b

Explanation:

7 0

3 years ago

PLEASE HELP! I don't get it at all! Speed is one thing; distance is another. Where is the arrow you shoot up at 50m/s when it ru

LuckyWell [14K]

I got you b, V(final)^2=V(initial+2acceleration*displacement
So this turns to (0m/s)^2=(50m/s)^2+2(9.8)(d) so just flip it all around to isolate d so you get
-(50m/s)^2/2(9.8) = d so you get roughly 12.7555 meters up

4 0

3 years ago

Read 2 more answers

A mass resting on a horizontal, frictionless surface is attached to one end of a spring; the other end is fixed to a wall. It ta

Tomtit [17]

Answer:

(a) 185 N/m

(b) 3.083 kg

Explanation:

(a)

Using,

E = 1/2ke²....................... Equation 1

Where E = work done to compress the spring, k = spring constant of the spring, e = compression of the spring.

make k the subject of the equation

k = 2E/e²............... Equation 2

Given: E = 3.7 J, e = 0.20 m

Substitute into equation 2

k = 2(3.7)/0.2²

k = 185 N/m.

(b)

Using,

F = ma.............. Equation 2

Where F = force applied to the spring, m = mass attached to the spring, a = acceleration of the spring.

But from hook's law,

F = ke................. Equation 3

substitute equation 3 into equation 2

ke = ma

make m the subject of the equation

m = ke/a................ Equation 4

Given: k = 185 N/m, e = 0.2 m, a = 12 m/s²

Substitute into equation 4

m = 185(0.2)/12

m = 3.083 kg

3 0

3 years ago

Read 2 more answers