Question

A satellite of mass 66 kg is in orbit round the Earth at a distance of 5.7R above its surface, where R is the value of the mean. radius of the Earth. If the gravitational field strength at the Earth's surface is 9.8 N kg ¹, calculate the centripetal force acting on the satellite. Assuming the Earth's mean radius to be 6400 km, calculate the period of the satellite in orbit in hours. ​

Answer 1

Answer: 14.4N , 24.5hrs.

Step-by-step explanation:

Answer 2

Answer:5.467

Explanation:The gravitational force between two objects is given by Newton's law of universal gravitation:

�

=

�

⋅

�

1

⋅

�

2

�

2

F=

r

2

G⋅m

1

​

⋅m

2

​

​

where:

�

F = gravitational force

�

G = gravitational constant (

6.67430

×

1

0

−

11

 

m

3

 

kg

−

1

 

s

−

2

6.67430×10

−11

m

3

kg

−1

s

−2

)

�

1

m

1

​

= mass of the Earth (

5.972

×

1

0

24

 

kg

5.972×10

24

kg)

�

2

m

2

​

= mass of the satellite (

66

 

kg

66kg)

�

r = distance between the center of the Earth and the satellite (given as

5.7

�

5.7R)

We need to express the distance

�

r in meters, so we first need to find the value of

�

R, the mean radius of the Earth, in meters:

Given mean radius of the Earth,

�

=

6400

 

km

=

6400

×

1000

 

m

=

6

,

400

,

000

 

m

R=6400km=6400×1000m=6,400,000m

Now, the distance

�

r between the center of the Earth and the satellite is:

�

=

5.7

�

=

5.7

×

6

,

400

,

000

 

m

r=5.7R=5.7×6,400,000m

Now, let's calculate the centripetal force

�

�

F

c

​

acting on the satellite:

�

�

=

�

⋅

�

1

⋅

�

2

�

2

F

c

​

=

r

2

G⋅m

1

​

⋅m

2

​

​

�

�

=

6.67430

×

1

0

−

11

×

5.972

×

1

0

24

×

66

(

5.7

×

6

,

400

,

000

)

2

F

c

​

=

(5.7×6,400,000)

2

6.67430×10

−11

×5.972×10

24

×66

​

�

�

≈

2.664

×

1

0

4

 

N

F

c

​

≈2.664×10

4

N

The centripetal force acting on the satellite is approximately

2.664

×

1

0

4

 

N

2.664×10

4

N.

Now, to calculate the period of the satellite's orbit, we can use Kepler's third law, which relates the orbital period (

�

T) of a satellite to the orbital radius (

�

r) and the mass of the Earth (

�

1

m

1

​

):

�

=

2

�

�

3

�

⋅

�

1

T=2π

G⋅m

1

​

r

3

​

​

We already have the values for

�

r and

�

⋅

�

1

G⋅m

1

​

:

T = 2 \pi \sqrt{\frac{{(5.7 \times 6,400,000)^3}}{{G \cdot 5.972 \times 10^{24}}}

Now, let's calculate the period

�

T:

�

≈

2

�

(

5.7

×

6

,

400

,

000

)

3

6.67430

×

1

0

−

11

×

5.972

×

1

0

24

T≈2π

6.67430×10

−11

×5.972×10

24

(5.7×6,400,000)

3

​

​

�

≈

1.968

×

1

0

4

 

seconds

T≈1.968×10

4

seconds

Finally, to express the period in hours, divide by the number of seconds in an hour:

�

≈

1.968

×

1

0

4

3600

≈

5.467

 

hours

T≈

3600

1.968×10

4

​

≈5.467hours

The period of the satellite in orbit is approximately 5.467 hours.