Question

$\huge \sf\fbox \gray{Question}$

1) If mass of a planet is it time the mass of the earth and its radius is twice the radius of the earth what will be the escape velocity for that planet.

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tiny{answer \: = 22.4km}$
$\underline{ \rule{190pt}{2pt}}$
2) how much time is satellite in an orbit at height 35780 km above Earth's surface will take if the mass of the earth would have been four times its original mass.

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tiny{answer \: = \: 12 \: hrs}$
$\underline{ \rule{190pt}{2pt}}$

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​

Answer 1

Answer:

Ans no 1

Solution

Given Mass of planet (Mp) ltbrge 8M (mass of earth),

radius of planet (Rp)

=2R (radius of earth )

To find : Escape velocity of planet (vp)

Formual : vesc=2GMR−−−−−√

Calculation : From formula , vp=2GmpRp−−−−−−√=2G(8M)2R−−−−−−−−√

=82×(2GMR)−−−−−−−−−−−−√

4×(2GMR)−−−−−−−−−−−√

but , (2GMR) is the escape velocity of the earth (vesc) .

∴ vp=2vesc

We know that , vesc =11.2 km/s

∴ vp=2×11.2=22.4km/s

The escape velocity for the planet will be 22.4 km/s .

Ans no 2

Solution

Given : New Mass of Earth (M) =4×(6×1024)kg

=24×1024kg.

Altitude of the satellite (h) =35780km .

=35780×103m

To find : Time required for revolution . (T) = ?

Formulae : V=GMR+h−−−−−−√

T=2π(R+h)v

Solution : V=GMR+h−−−−−−√

∴V=6.67×10−11×24×10246.4×106+35780×103−−−−−−−−−−−−−−−−−−−−√

∴V=37.59×106−−−−−−−−−√

V=6.160km/s

T=2π(R+h)v

=2×3.14×6400+357806.16

=43001.68sec

T=11.94hr~12 hours .

∴ The satellite will take approximately 12 hours to revolve around earth.

Hope it helps

Answer 2

Answer:

1) Given in the question :-

Planet mass is 8 times that earth

Planet radius is twice then the radius of earth.

Let mass of the Earth = M

Radius of the Earth = R

Now, we know that The equation to get escape velocity

$v = √(2 g r)$

If the mass of planet is eight times then escape velocity = 2 √2 times.

Escape velocity reduced to by 1/√2. If radius expands twice

Hence the net escape velocity will be increase.

$2 √2 × (1/√2)=$ 2 times now.

Now it is clear that the escape velocity of that planet will be 2 times from the earth $22.4 km/s$

2) Given:

Height of the satellite, h=35780 km

Let the original mass of Earth be M. Then its bew mass will be 4M.

Time taken by the satellite to revolved around the Earth's orbit is given as

T=

v

c

​

2πR

​

Now, v

C

​

 is given as

v

c

​

=

R+h

GM

​

​

Thus,

T=

R+h

GM

​

​

2π(R+h)

​

=

GM

​

2π(R+h)

R+h

​

​

 .....(i)

⇒T∝

M

​

1

​

 ... (ii)

Thus from equation (ii) we see that when the mass of the Earth becomes

4 times, the time period of revolution of satellite should be halved.

i.e. T

4M

​

=

2

T

​

 ....(iii)

Now, h=35780 km

M=6×10

24

kg

R=6.4×10

5

m

G=6.67×10

−11

N m

2

/kg

2

Putting the values of h,M and R in first, we get

T=24 h

Using (iii), we get

T

4M

​

=

2

24

​

=12 hGiven:

Height of the satellite, h=35780 km

Let the original mass of Earth be M. Then its bew mass will be 4M.

Time taken by the satellite to revolved around the Earth's orbit is given as

T=

v

c

​

2πR

​

Now, v

C

​

 is given as

v

c

​

=

R+h

GM

​

​

Thus,

T=

R+h

GM

​

​

2π(R+h)

​

=

GM

​

2π(R+h)

R+h

​

​

 .....(i)

⇒T∝

M

​

1

​

 ... (ii)

Thus from equation (ii) we see that when the mass of the Earth becomes

4 times, the time period of revolution of satellite should be halved.

$i.e. T 4M​ = 2T​ ....(iii)Now, h=35780 km$

$24/ 2$

​ $=12h$

Explanation:

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