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Kryger [21]
3 years ago
14

A satellite in orbit around the Earth has a speed of 8 km/s at a given point of its orbit. If the period is 2 h, what is the alt

itude at that point?
Physics
1 answer:
zzz [600]3 years ago
7 0

Answer:

Explanation:

Using the formula;

Speed = Distance/Time

Given

Speed = 8km/s

Time = 2hr

Convert 2hr to secs

1hr = 3600secs

2hr = 2(3600)

2hr = 7200secs

Altitude is the distance

From the formula;

Distance = speed × time

Distance = 8×7200

Distance = 57200km

Hence the altitude at this point is 57,200km

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A 545-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli
Art [367]

Answer

given,

mass of satellite = 545 Kg

R = 6.4 x 10⁶ m

H = 2 x 6.4 x 10⁶ m

Mass of earth = 5.972 x 10²⁴ Kg

height above earth is equal to earth's mean radius

a) satellite's orbital velocity

   centripetal force acting on satellite = \dfrac{mv^2}{r}

     gravitational force = \dfrac{GMm}{r^2}

    equating both the above equation

    \dfrac{mv^2}{r} = \dfrac{GMm}{r^2}

      v = \sqrt{\dfrac{GM}{r}}

      v = \sqrt{\dfrac{6.67 \times 10^{-11}\times 5.972 \times 10^{24}}{2 \times 6.4 \times 10^6}}

          v = 5578.5 m/s

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

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

          T = 14416.92 s

          T = \dfrac{14416.92}{3600}\ hr

          T = 4 hr

c) gravitational force acting

  F = \dfrac{GMm}{r^2}

  F = \dfrac{6.67 \times 10^{-11}\times 545 \times 5.972 \times 10^{24} }{(6.46 \times 10^6)^2}

     F = 5202 N

4 0
3 years ago
A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit
Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M=8.7\cdot 10^{13}kg is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

m\frac{v^2}{r}=G\frac{Mm}{r^2}

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s

3 0
3 years ago
What distance will a truck trvael in 3 hours at an average speed of 50 per hour​
FromTheMoon [43]
The truck would of went 150 miles
5 0
2 years ago
A sailboat picks up a gust of wind and accelerates to a speed of 6m/s in 16 seconds.If the initial velocity 1 m/s,what is the ac
noname [10]

Answer:

0.3125m/s

Explanation:

3 0
3 years ago
What is an unbalanced force
Butoxors [25]
There's no such thing as "an unbalanced force".

If all of the forces acting on an object all add up to zero, then we say that
<span>the group </span>of forces is balanced.  When that happens, the group of forces
has the same effect on the object as if there were no forces on it at all. 

An example: 
Two people with exactly equal strength are having a tug-of-war.  They pull
with equal force in opposite directions.  Each person is sweating and straining,
grunting and groaning, and exerting tremendous force.  But their forces add up
to zero, and the rope goes nowhere.  The <u>group</u> of forces on the rope is balanced.

On the other hand, if one of the offensive linemen is pulling on one end of
the rope, and one of the cheerleaders is pulling on the other end, then their
forces don't add up to zero, because even though they're opposite, they're
not equal.  The <u>group</u> of forces is <u>unbalanced</u>, and the rope moves.

A group of forces is either balanced or unbalanced.  A single force isn't.
7 0
3 years ago
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