Question

BRAINLIST!!! ANSWER PLEASE •_• a spaceship orbiting around the earth at a distance 5 of times the radius of the earth from its Centre if the acceleration due to gravity on the earth is 9.8 m s^-2 calculate the acceleration due to gravity acting on the spaceship.​

Answer 1

Answer:

Explanation:

Given that,

Acceleration a=2.45m/s  

2

 

Mass of earth M=6×10  

24

kg

Radius of earth R=6×10  

6

m

Gravitational constant G=6.67×10  

−11

Nm  

2

/kg  

Now centripetal acceleration is

 a  

c

​  

=  

R

v  

2

 

​  

 

a  

c

​  

=  

R

(  

R

GM

​  

 

​  

)  

2

 

​  

 

a  

c

​  

=  

R  

2

 

GM

​  

 

a  

c

​  

=  

(R+h)  

2

 

GM

​  

 

Now, put the values

 2.45=  

(6×10  

6

+h)  

2

 

6.67×10  

−11

×6×10  

24

 

​  

 

(6×10  

6

+h)  

2

=  

2.45

6.67×10  

−11

×6×10  

24

 

​  

 

(6×10  

6

+h)  

2

=  

2.45

40.02×10  

13

 

​  

 

(6×10  

6

+h)  

2

=1.6334×10  

14

 

6×10  

6

+h=  

1.6334×10  

14

 

​  

 

h=1.278×10  

7

−6×10  

6

 

h=(12.78−6)×10  

6

 

h=6.78×10  

6

 

h=6.8×10  

6

m

So, h≅R

Hence, the height of spaceship is R