(Q1)At what altitude above the earth's surface would the acceleration due to gravity be 4.9m/s-2?Assuming the mean radius of the earth is 64000000meters and the acceleration due to gravity 9.8m/s-2 on the surface of the earth (a)26000000m(b)323000000m(c)465000000m(d)776000000m(Q2)A man weighs 750 N on the surface of the earth.What will be his weight when standing on the moon? The masses of the earth and the moon are respectively 59800000000kg and 73600000 kg.Their radii are respectvely 637000km and 174000km (a)200.5N(b)123.7(c)550.4N(d)1000.0N.(Q3)A 2000kg satelite orbits the earth at a height of 300km. What is the speed of the satellite and its period?Take G=0.0000000000667Nm2/kg2 Mass of earth is 598000000kg (a)7.73 km/s 54000(b)855.4km/s and 770000(c)497.2km/s 5500000(d)322.3km/s and 430000

Q1

You want to have 1/2 the gravitational acceleration at the surface.
Since F ∝ 1/r^2, you will need the new height to be √2 times the surface radius. That is, at an altitude of (√2-1)R

(√2-1) * 64000000 = 26500000

so I'd choose (a)

Use similar logic on the others.

Q1

the gravitational constant G =6.67•10⁻¹¹ N•m²/kg²,
Radius of the Earth is 6400 000 m =6.4•10⁶ m ===Your data is incorrect!!!

mg=G•m•M/R²
mg`= G•m•M/(R+h)²
(R+h)²/ R² = g /g`=9.8/4.9 = 2
R²+2Rh + h²= 2R²
h²+2R - R²= 0
h = - R ±sqrt(2R²) = - R ± R√2 =>
h=R(√2 -1) = 0.41 R=
=0.41 •6.4•10⁶ =2624000 m
Ans. (a)
Q2
Mass of the Earth = 5.98•10²⁴ kg,
Radius of the Earth = 6.4•10⁶ m,
Mass of the Moon = 7.36•10² kg,
Radius of the Moon = 1.74•10⁶ m,

mg= G•m•M/R²,
mg`= G•m•M`/R`²,
g/g`= M R`²/R² M`,
g`=g R² M`/ M R`²=
=9.8•7.36•10²²(6.4•10⁶)²/5.98•10²⁴(1.74•10⁶)² = =1.63.

W=mg => m =W/g=750/9.8 =76.53 kg
mg`= 76.53•1.63 = 124 N

Q3

mv²/(R+h) = G•m•M/(R+h)²,
v=sqrt{GM/(R+h)} =
=sqrt{6.67•10⁻¹¹•5.98•10²⁴/(6.4•10⁶ +3•10⁵) =
= 7730 m/s =7.73 km/s.

T=2π(R+h)/v =
= 2π(6.4•10⁶ +3•10⁵)/4730 ≈ 5400 s.

(Q1) To find the altitude at which the acceleration due to gravity is 4.9 m/s^2, we can use the formula for gravitational acceleration:

g = (G * m) / r^2

Where:
g = gravitational acceleration
G = gravitational constant (6.67 x 10^-11 Nm^2/kg^2)
m = mass of the Earth (5.98 x 10^24 kg)
r = radius from the center of the Earth

Let's solve for the altitude:

First, we need to find the value of g at the surface of the Earth:
g = 9.8 m/s^2

Next, we can set up the following equation:

4.9 = (G * m) / (r + h)^2

Where h is the altitude we're trying to find.

Since we know the value of g at the surface of the Earth, we can substitute all the known values:

4.9 = (6.67 x 10^-11 Nm^2/kg^2 * 5.98 x 10^24 kg) / (6400000 + h)^2

Now we can solve for h:

4.9 * (6400000 + h)^2 = 6.67 x 10^-11 Nm^2/kg^2 * 5.98 x 10^24 kg

(6400000 + h)^2 = (6.67 x 10^-11 Nm^2/kg^2 * 5.98 x 10^24 kg) / 4.9

(6400000 + h)^2 = 5.12266256757 x 10^12

Taking the square root of both sides:

6400000 + h = sqrt(5.12266256757 x 10^12)

h = sqrt(5.12266256757 x 10^12) - 6400000

Calculating the value on a calculator:

h ≈ 26000000 meters

Therefore, the correct option is (a) 26000000 meters.

(Q2) To find the man's weight on the Moon, we can use the equation for gravitational force:

F = (G * m1 * m2) / r^2

Where:
F = gravitational force
G = gravitational constant (6.67 x 10^-11 Nm^2/kg^2)
m1 = mass of the man
m2 = mass of the Moon
r = distance between the man and the Moon's center

We need to find the weight, which is the gravitational force acting on the man.

Weight = m * g

To find the weight on the Moon, we need to calculate the gravitational force on the surface of the Moon using the given masses and radii.

First, let's find the gravitational force on the surface of the Earth:

Weight = m * g = 750 N (Given)

Now let's calculate the gravitational force on the Moon's surface:

750 N = (G * m1 * m2) / (637000000 + 174000)^2

Using the given values:

750 N = (6.67 x 10^-11 Nm^2/kg^2 * m1 * 73600000 kg) / (81100000000)^2

Now, let's solve for m1 (man's mass):

m1 = (750 N * (81100000000)^2) / (6.67 x 10^-11 Nm^2/kg^2 * 73600000 kg)

Calculating the value on a calculator:

m1 ≈ 76.967 kg

Now we can calculate the weight on the Moon:

Weight = m * g = 76.967 kg * 1.62 m/s^2 (Acceleration due to gravity on the Moon is 1.62 m/s^2)

Weight ≈ 124.680 N

Therefore, the correct option is (b) 123.7 N.

(Q3) To find the speed of the satellite, we can use the formula for orbital velocity:

v = sqrt((G * M) / r)

Where:
v = orbital velocity
G = gravitational constant (6.67 x 10^-11 Nm^2/kg^2)
M = mass of the Earth (5.98 x 10^24 kg)
r = distance from the center of the Earth

Let's find the speed:

First, let's calculate the altitude from the height given:
Altitude = height - radius of the Earth = 300000 - 6400000 = -6100000 m

Now, let's calculate the speed:

v = sqrt((6.67 x 10^-11 Nm^2/kg^2 * 5.98 x 10^24 kg) / (6400000 + -6100000) m)

Calculating the value on a calculator:

v ≈ 8554 m/s

Therefore, the speed of the satellite is approximately 8554 m/s.

To find the period, we can use the formula for the orbital period:

T = (2 * pi * r) / v

Where:
T = orbital period
pi = 3.14159
r = distance from the center of the Earth (6400000 + -6100000) m
v = orbital velocity (8554 m/s)

Let's calculate the period:

T = (2 * 3.14159 * (6400000 + -6100000) m) / 8554 m/s

Calculating the value on a calculator:

T ≈ 77000 s

Therefore, the period of the satellite is approximately 77000 seconds.

Therefore, the correct option is (c) 855.4 km/s and 770000 seconds.

To find the answers to these questions, we need to apply the principles of gravity and the laws of motion. Let's go through each question one by one and explain the steps to find the answers.

Q1) At what altitude above the Earth's surface would the acceleration due to gravity be 4.9 m/s²?

The acceleration due to gravity decreases as you move away from the Earth's surface. To find the altitude where the acceleration is 4.9 m/s², we can use the formula for gravitational acceleration at a given altitude:

a = (G * M) / (R + h)²

Where:
a = acceleration due to gravity at altitude
G = gravitational constant (6.67 x 10^-11 Nm²/kg²)
M = mass of the Earth (5.98 x 10^24 kg)
R = mean radius of the Earth (6.4 x 10^6 meters)
h = altitude above the Earth's surface

Re-arranging the formula to solve for h, we get:

h = (G * M) / (a)^(1/2) - R

Now we substitute the known values:

a = 4.9 m/s²
G = 6.67 x 10^-11 Nm²/kg²
M = 5.98 x 10^24 kg
R = 6.4 x 10^6 meters

Calculating the value of h using the formula above will give us the altitude where the acceleration due to gravity is 4.9 m/s².

Q2) What will be a man's weight when standing on the moon if he weighs 750 N on the surface of the Earth?

To find the weight on the moon, we can use the formula for gravitational force:

F = (G * m1 * m2) / r²

Where:
F = gravitational force
G = gravitational constant (6.67 x 10^-11 Nm²/kg²)
m1 = mass of the man
m2 = mass of the moon
r = distance between the center of the moon and the man

Since weight is the force exerted by gravity on an object, we can use the formula to find the weight on the moon by substituting the known values:

F = 750 N (weight on the surface of the Earth)
G = 6.67 x 10^-11 Nm²/kg²
m1 = mass on the surface of the Earth
m2 = mass of the moon
r = distance from the center of the moon to the man's location on the moon's surface

Calculating the value of m1, m2, and r using the given mass and radii of the Earth and the Moon respectively, we can find the weight on the moon.

Q3) What is the speed of a satellite and its period if it orbits the Earth at a height of 300 km?

To find the speed and period of the satellite, we can use the principles of circular motion and the equation for centripetal force.

The centripetal force acting on the satellite is provided by gravitational force:

F = (G * m1 * m2) / r²

Where:
F = gravitational force (provided by centripetal force)
G = gravitational constant (6.67 x 10^-11 Nm²/kg²)
m1 = mass of the satellite
m2 = mass of the Earth
r = distance between the center of the Earth and the satellite's orbit

The centripetal force is also equal to the mass of the satellite multiplied by its acceleration, which is the centripetal acceleration:

F = m * a (centripetal force = mass * centripetal acceleration)

From the above two equations, we can equate them and solve for the satellite's centripetal acceleration:

(G * m1 * m2) / r² = m * a

Rearranging the formula to solve for a:

a = (G * m2) / r²

We can also calculate the velocity (v) of the satellite at a given radius (r) using the formula:

v = (G * m2 / r)^(1/2)

The period (T) of the satellite is the time it takes to complete one orbit and can be found using:

T = (2π * r) / v

Substituting the known values into the formulas, including G, m2, and r from the given values, we can find the speed and period of the satellite.