All categories

3 years ago

What is the kinetic energy of a 200 kg satellite as it follows a circular orbit of radius 8x106m around the earth?

Physics

2 answers:

Lerok [7]3 years ago

7 0

Answer:

Kinetic energy of satellite will be 4.98 × 10^9 J.

Explanation:

The velocity V of a satellite orbiting around the earth is given by

V = √(GM/r) ………. (i)

Where,

G = gravitational constant = 6.67 × 10^-11 Nm^2/kg^2

M = mass of earth = 5.97 × 10^24 kg

r = radius of orbit = 8 × 10^6 m

By putting values in equation (i),

V = √((6.67 × 10^-11)(5.97 × 10^24)/ 8 × 10^6)

V = 7055.13 m/s

We know that,

Kinetic energy = ½ mV^2

Where,

m = mass of satellite

so,

Kinetic energy = ½ (200)(7055.13)^2

Kinetic energy = 4.98 × 10^9 J

motikmotik3 years ago

4 0

Given the equation for the Speed of a Satellite

v = SqRt{Gravitational Constant}{Mass of Earth} divided by the radius given in your problem

we have:

(square root whole term on right side)

v = G Me
———
r

so. (6.67x10^-11)(5.97x10^24)
___________________
(8.0x10^6)

v = 7055 m/s (which is reasonable)

so utilize the Kinetic Energy Formula

KE = 1/2mv^2

KE = 1/2(200)(7055)^2

KE = 4.977x10^9 J

You might be interested in

Please help!<br>The question is attached!!

Law Incorporation [45]

Answer:

the pic is blur,btw can u type sweetie

7 0

3 years ago

Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance

siniylev [52]

Answer:

interest point:

1) Point on the left side

2) Point within the radius r₁ of the first sphere

3) Point between the two spheres

4) point within the radius r₂ of the second sphere

5) Right side point

Explanation:

In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres

We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is

E_ {total} = E₁ + E₂

E_{ total} = $k \frac{Q}{x_1^2} + k \frac{Q}{x_2^2}$

the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d

x₂ = x₁ -d

E total = $k \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}$

Let's analyze the field for various points of interest.

1) Point on the left side

in this case

E_ {total} = $k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )$

E_ {total} = $k \frac{Q}{x_1^2}$ $( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )$

We have several interesting possibilities:

* We can see that as the point is further away the field is more similar to the field created by two point charges

* there is a point where the field is zero

E_ {total} = 0

x₁² = (x₁ + d)²

2) Point within the radius r₁ of the first sphere.

In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point

E_ {total} = $-k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}$